Concept of Limit

Limit is a tool for studying differential and integral calculus. Derivatives, definite integrals and infinite series are all defined by limit. Different from elementary algebra, the concept of limit relies on a dynamic process of change, and thus calculus that is based on limit further studies rates and amount of changes for variables and functions.

Basic concept of limit Let \(f\) be a function defined for all \(x\) in an open interval containing \(c\), except possibly at \(x = c\), and let \(L\) be a real number. We say the limit of \(f(x)\) is \(L\) if \(f(x)\) is arbitrarily close to \(L\) as \(x\) approaches \(c\), and write \(f(x) → L\) as \(x → c\). Or sometimes say \(f(x)\) converges to \(L\) as \(x\) tends to \(c\).

This means if \(f(x)\) has a limit \(L\), then \(f(x)\) can be made as close to \(L\) as we want by taking values of \(x\) sufficiently close to \(c\). Note the following points.
(1) The value \(f(c)\) may or may not be defined, but it does not matter. All matters is the values of \(f(x)\) are defined for all \(x\) near \(c\).
(2) If \(f(x)\) does not converges to any number as \(x\) approaches \(c\), the limit of \(f(x)\) does not exist.
(3) If the limit of \(f(x)\) exists, then it is unique.
(4) The words "\(f(x) → L\) as \(x → c\)" means we can make the value \(f(x)\) as close to \(L\) as we want merely by choosing a corresponding number \(x\) sufficiently close to \(c\).
(5) The expression "\(f(x)\) is arbitrarily close to \(L\)" are not precise, because the meaning of how close is close depends on specific contexts. A more precise description for closeness is the distance \(|f(x) - L|\), which can be made smaller than any number \(ε > 0\) we choose. So we use distance in the formal definition of limit.

Finite and infinite limits If \(L\) is the resulting limit of \(f(x)\) as \(x\) approaches \(c\) and \(L\) is a real number, we call \(L\) a finite limit. If \(L\) is infinity either \(\infty\) or \(-\infty\), we call it an infinite limit. In this case, \(f(x)\) is arbitrarily large or small. Or as \(x\) approaches \(c,f(x)\) increases or decreases without bound.

One-sided limit If \(L\) is the resulting limit of \(f(x)\) as \(x\) approaches \(c\) from the left side of \(c\), we call \(L\) the left-side limit. Similarly, we call \(L\) the right-side limit if it is the limit of \(f(x)\) as \(x\) approaches \(c\) from the right side.

Left- or right-side limit is also called one-sided limit. The function \(f\) has a limit \(L\) as \(x\) approaches \(c\) if and only if both left-side and right-side limits exist and are equivalent.

Limit at infinity If \(L\) is the resulting limit of \(f(x)\) as \(x\) tends to infinity, we call \(L\) the limit at infinity. Limits at infinity describe the asymptotic behaviors of functions as \(x\) increases or decreasing without bound, so we usually call the line \(y = L\) a horizontal asymptote to the graph of \(y = f(x)\).

Limit does not exist The function \(f\) does not have to have a limit as \(x\to c\). If \(f(x)\) does not converge to either a finite number or tends to infinity as \(x\to c\), we say the limit does not exist. There are various situations where limits fail to exist.

(1) Function undefined near c If the limit of \(f(x)\) exists as \(x \to c, f(x)\) should be defined for all \(x\) in an open interval containing \(c\). The limit \(\lim\limits_{x\to-3}\sqrt{x}\) does not exist because the function is not defined on any interval near -3, and \(\lim\limits_{x\to 0}\sqrt{x}\) does not exist, either, because there is no open interval containing 0 in the domain \(x≥0\). For similar reasons, the limits \(\displaystyle\lim_{x\to0}\ln x,\lim_{x\to 0}x^{\frac{3}{2}}\) do not exist.

(2) Oscillation If values of \(f(x)\) oscillates as \(x \to c\), the limit does not exist. The limit \(\displaystyle\lim_{x\to 0}\sin\frac{1}{x}\) does not exist because \(\sin \frac{1}{x}\) oscillates from -1 to 1 as \(x\) approaches 0. These limits \(\displaystyle\lim_{x\to\pm \infty}\cos x,\lim_{n\to\infty}(-1)^n\) do not exist.

(3) Vertical asymptotes The limits \(\displaystyle\lim_{x\to\pi}\cot x,\lim_{x\to0}\frac{1}{x}\) do not exist because \(\cot(x)\) has a vertical asymptote at \(x = π\) and \(\frac{1}{x}\) has a vertical asymptote at \(x = 0\). Similarly, \(\displaystyle\lim_{x\to\frac{\pi}{2}}\tan x, \lim_{x\to1}\ln(x-1),\lim_{x\to2}\frac{1}{\sqrt{x^2-4}}\) do not exist.

(4) Inequivalent two-sided limits If two-sided limits exist but are not equivalent, the limit fails to exists. Since \(\displaystyle\lim_{x\to1^+}⌊x⌋=1,\lim_{x\to1^-}⌊x⌋=0,\lim_{x\to1^-}⌊x⌋\) does not exist. The limit \(\lim\limits_{x\to 0}\frac{|x|}{x}\) does not exist either, because the limit is 1 as \(x\) approaches 0⁺, and is -1 as \(x\) approaches 0^-.

Find limit of f(x): [lim; f(x); x; c; n/p/np], [lim(f(x), x, c, n/p/np)]

To find the limit of function f(x) as x approaches c, you need to enter at least four items "lim; f(x); x; c" in this order, where 'lim' is the operation code, 'f(x)' represents the function expression, 'x' the independent variable, and 'c' the number that 'x' approaches.

For one-sided limits, you need to add the keyword 'n' (negative '-' or left side), and 'p' (positive '+' or right side) to the end. The default limit is two-sided. Find limits of single variable functions, either one-sided or two-sided. Multivariate limits are not supported.

Attention Write "lim(f(x), x, c, n/p) to find limits of f(x) as x approaches a value "c", and you get the same results as "lim; f(x); x; c; n/p".

Examples
(1) The code lim;2*x;x;0 || find the limit of \(f(x) = 2x\) as \(x\to 0\).
(2) lim;1/(x+2);x;0 || lim;x**(1/2);x;4 || lim;x**(1/2);x;-3 || lim;1/x;x;0 || lim;abs(x)/x;x;0 || lim;floor(x);x;1 || lim;tan(x);x;pi/2 || lim;log(x-1);x;1 || lim;1/(x**2-4)**(1/2);x;2 || lim;x**(3/2);x;0;p || lim;log(x);x;p || lim;(1-x)**(1/2);x;1;n || lim;1/(x-3);x;3;n || lim;1/x;x;0;n || lim;abs(x)/x;x;n || lim;(a+2)/(2-a);a;2;p ||.
(3) Check lim((x**3-1)/(x-1),x,1) || lim((a-4)/(a**(1/2)-2),a,4,n) || lim(sin(1/x),x,oo) ||.

Symbols for \(\infty,\pi,e\) To find limit at infinity, you need to enter 'oo' for infinity \(\infty\) and '-oo' for \(-\infty\). Similarly, enter "pi" for the number \(\pi\), and "e" or "E" for \(e\).

Examples lim;sin(1/x);x;oo || lim;cos(x);x;-oo || lim;(-1)**n;n;oo || lim;cot(x);x;pi || lim;1/x;x;-oo || lim;log(x);x;e ||
lim;(1+1/n)**n;n;oo || lim;2-5*x;x;-oo || lim;8*x-3;x;oo || lim;(x+4)/(x**2-3*x+1);x;oo || lim;(a**(1/2)-3)/(a+5);a;oo ||
lim;(2-4*b)/(b**2-2*b+5)**(1/2);b;oo || lim;(d+7)**(1/2)-d**(1/2);d;oo || lim;(z**2+z)**(1/2)-z;z;oo || lim;cos(x);x;oo || lim;(-1)**n;n;oo ||.

Limit as a complex number You may notice that limits may be complex numbers. If you only consider functions and limits of real numbers, a complex limit can be regarded as the limit does not exist.

Examples
(1) The limit \(\lim\limits_{x\to-1}\sqrt{x}\) does not exist in real values, so lim;x**(1/2);x;-1 || gives the limit \(i\). Since a square root function is defined on nonnegative numbers only, the two-side limit \(\lim\limits_{x\to 0}\sqrt{x}\) does not exist. In complex numbers, however, the limits lim;x**(1/2);x;0 || lim;x**(1/2);x;0;n || exist.
(2) Similar results by lim;log(x);x;0 || lim;log(x);x;0;p || lim;log(x);x;0;n || lim;log(x);x;oo || lim;log(x);x;-oo ||.

Examples lim;sin(x);x;oo || lim;sin(x)/x;x;0 || lim;(x**2-4)/(x-2);x;2 || lim;floor(x);x;3 || lim;floor(x);x;-5;p || lim;ceiling(x);x;-1.5 || lim;ceiling(t);t;0 || lim;1/x;x;-oo || lim;abs(x)/x;x;0 || lim;(x**2-x-6)/(x-3);x;3 || lim;((x+5)**(1/2)-5**(1/2))/x;x;0 || lim;1/(x-2)**2;x;2 || lim;(x+4)/(x-1);x;1 || lim;(1+n)**(1/n);n;0 || lim;(y+1/y)**y;y;oo || lim;(1-2/u)**u;u;oo || lim;tan(t)-sec(t);t;pi/2 || lim;cot(x)*csc(x);x;0 || lim;csc(x)-1/x;x;0 || lim;cos(5*x)/cos(3*x);x;pi/2 || lim;x**(cos(x));x;0;p || lim;log(1+u)/u;u;0 || lim;log(1+4*t)/tan(2*t);t;0 || lim;(exp(y)-1)/sin(y);y;0 || lim;(w**5-1)/(w-1);w;1 || lim;(1+2**x)**(1/x);x;oo || lim;(1-cos(z))/z**4;z;0 || lim;cos(t)**(1/abs(t));t;0 || lim;((1+h)**(1/3)-1)/h;h;0 || lim;((1+h)**(1/3)-1)/((1+h)**(1/2)-1);h;0 || lim;(5*x**3-x)/(2*x**2+3)*sin(1/x);x;oo.

Formal Definition of Limit: [lim; f(x); x; c; n/p; ε], [lim(f(x), x, c, n/p, ε) ]

Finite limits A function \(f\) has a finite limit \(L\) as \(x\) approaches \(c\) means for any \(\epsilon > 0\) there is a corresponding \(\delta > 0\) such that if \(0<|x-c|<\delta\), then \(|f(x)-L| < \epsilon\).

In other words, if \(L\) is a real number, no matter how small \(\epsilon\) is chosen, we can always find a corresponding \(\delta\) such that \(x∈(c-\delta,c+\delta)\) implies \(f(x)∈(L-ε,L+ε)\). Or if \(x\) is within \(\delta\) of \(c, f(x)\) is within \(\epsilon\) of \(L\).

Note these important points in the formal definition.
(1) The inequality \(0 < |x-c| < \delta\) excludes the point \(x=c\), which means the limit of \(f(x)\) depends only on the values of \(f(x)\) for \(x\) near \(c\) but not equal to \(c\).
(2) If the limit \(L\) of \(f(x)\) exists as \(x\) approaches \(c\), then \(L\) is unique.
(3) This \(\epsilon-\delta\) definition is not used to find limits, but used to prove limits and their properties.
(4) To prove limits is usually to relate the inequality \(|f(x)-L| < \epsilon\) to \(0 < |x-c| < \delta\) and find \(\delta\).
(5) The choice of \(\delta\) depends on the choice of \(\epsilon\). This means if the limit exists, then for each choice \(\epsilon>0\), we can always find a \(\delta>0\) associated with the number \(\epsilon\) such that \(|f(x)-L|< \epsilon\) under the condition \(0 <|x-c|<\delta\).

Attention These expressions \(\displaystyle\lim_{x\to c}f(x)=L⇔\lim_{x\to c}[f(x)-L]=0⇔\lim_{x\to c}|f(x)-L|=0⇔\lim_{h\to 0}f(c+h)=L\) are equivalent.

Verify limits by formal definition, and find a corresponding \(\delta\) interval for any given \(ε\) value.

Attention Use the array "lim; f(x); x; c; n/p; ε" or "lim(f(x), x, c, n/p, ε)" for finding a δ interval for x near c. For two-sided limits, use "lim; f(x); x; c; ε" or "lim(f(x), x, c, ε)".

Examples
(1) The limit of \(f(x)=x²\) as \(x \to\) 2 is 4. Find the corresponding \(δ\) interval satisfying \(|x² - 4| < ε\) for arbitrary small number \(δ\) = 0.1, 0.05, 0.001, and 0.0001 ... by lim;x**2;x;2;0.1 || lim;x**2;x;2;0.05 || lim;x**2;x;2;0.001 || lim;x**2;x;2;0.0001 || lim(x**2,x,2,0.00001) ||.
(2) lim;1/(x-1);x;3;0.01 || lim;1/(x-1);x;3;0.001 || lim;1/(x-1);x;3;0.0005 || lim;log(1+u);u;0;0.01 || lim;t**(1/2);t;4;0.05 || lim;(z**2-9)/(z-3);z;3;0.001 || lim((z**2-9)/(z-3),z,3,0.001) ||.

Infinite limits A function \(f\) has an infinite limit \(\displaystyle\lim_{x\to c}f(x)=\infty\) if for every \(M>0\), there exists a corresponding \(\delta > 0\) such that \(f(x) > M\) whenever \(0<|x-c|<\delta\).

A function \(f\) has an infinite limit \(\displaystyle\lim_{x\to c}f(x)=-\infty\) if for every \(N< 0\), there exist a corresponding \(\delta > 0\) such that \(f(x) < N\) whenever \(0< |x-c| < \delta\).

This definition implies no matter how large a positive number \(M\) is chosen, we can always find the corresponding \(\delta\) interval such that \(f(x)>M\) whenever \(x\in(c-\delta,c+\delta)\). Or no matter how small a negative number \(N\) is chosen, we can always find the corresponding \(\delta\) interval such that \(f(x)< N\) whenever \(x\in(c-\delta,c+\delta)\).

Examples
(1) The limit of \(y=\frac{1}{x}\) as \(x \to\) 0⁺ is positive infinity, which means no matter how large a positive number \(M\) we choose, we can always find the corresponding \(δ\) interval such that \(y > M\) whenever \(x\) in such an open set. Check \(M\) = 1000, 10000, 1000000, ..., by lim;1/x;x;0;p;1000 || lim;1/x;x;0;p;10000 || lim;1/x;x;0;p;1000000 ||.
(2) Check similar results as x approaches 0^- by lim;1/x;x;0;n;-500 || lim;1/x;x;0;n;-50000 || lim;1/x;x;0;n;-8000000 ||.
(3) lim;u**2;u;oo;p;500 || lim;log(w);w;0;p;-600 || lim;y**(1/2);y;oo;p;1000 || lim;2/(3-z);z;3;p;-500 || lim(2/(3-z),z,3,p,-500) || lim(y**(1/2),y,oo,p,1000) ||.

Attention The line \(x=c\) is called a vertical asymptote to the graph of \(y=f(x)\) if \(\displaystyle\lim_{x\to c^+}f(x)=\pm \infty\) or \(\displaystyle\lim_{x\to c^-}f(x)=\pm \infty\) or both.

Limit at infinity The limit \(\displaystyle\lim_{x\to\infty}f(x)=L\) means for any \(\epsilon >0\), there exist a corresponding \(M>0\) such that whenever \(x > M, |f(x)-L|< \epsilon\).

The limit \(\displaystyle\lim_{x\to -\infty}f(x)=L\) means for any \(\epsilon >0\), there exist a corresponding \(N< 0\) such that whenever \(x < N, |f(x)-L|< \epsilon\).

Examples The limit of 2^-n as n tends to infinity is 0, which implies no matter how small ε is chosen, we can always find M such that 2^-n < ε whenever n > M. Check if ε = 0.01, 0.0001, 0.000001 by lim;1/n**2;n;oo;p;0.001 || lim;1/2**n;n;oo;p;0.0001 || lim;1/2**n;n;oo;p;0.000001 ||. lim;1/x;x;oo;0.02 || lim;1/x;x;-oo;0.001 || lim;1/n;n;oo;p;0.001 || lim(1/n,n,oo,p,0.0001) ||.

One-sided limits The function \(f\) has a left-hand limit \(\displaystyle\lim_{x\to c^-}f(x)=L\) means for each \(\epsilon > 0\), there exists a corresponding \(\delta >0\) such that if \(0 < c-x <\delta\), then \(|f(x)-L|< \epsilon \).

The function \(f\) has a right-hand limit \(\displaystyle\lim_{x\to c^+}f(x)=L\) means for each \(\epsilon > 0\), there exists a corresponding \(\delta >0\) such that if \(0 < x-c <\delta\), then \(|f(x)-L|< \epsilon \).

The limit of \(f(x)\) as \(x\) approaches \(c\) exists if and only if both one-sided limits exist and are equivalent, or \(\displaystyle\lim_{x\to c}f(x)=L ⇔ \lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=L\).

Examples The limit of \(\frac{|x|}{x}\) as \(x \to\) 0⁺ is 1, which means for any \(ε > 0\), we can always find a correspond \(δ\) interval such that \(|\frac{|x|}{x} - 1| < ε\) if \(x\) is within such interval. Check if \(ε\) = 0.01, 0.0001, ... by lim;abs(x)/x;x;0;p;0.01 || lim;abs(x)/x;x;0;p;0.0001 ||. The limit is -1 as \(x \to\) 0⁺ by lim;abs(x)/x;x;0;n;0.01 || lim;abs(x)/x;x;0;n;0.0001 ||.

Limits of infinite sequences: [lim; f(n); n; oo]

An infinite sequence of numbers can be viewed as the values taken by a function \(f(n)\) whose domain is the set of whole numbers, so the definition of limit at infinity applies to sequence limit. To determine if a sequence \(a^n\) converges, type its general term \(a^n\), which is a function of \(n\), followed by \(n\) and "oo".

Examples
(1) The limit of an infinite sequence \(a^n= (1 + \frac{1}{n})^n\) is \(e\) by lim;(1+1/n)**n;n;oo || lim((1+1/n)**n,n,oo) ||.
(2) The limit \(a_n = \frac{\ln n}{n}\) is 0 by lim;log(n)/n;n;oo || lim(log(n)/n,n,oo) ||.
(3) The limit of \(n³e^{-n}\) is 0 by lim;n**3*exp(-n);n;oo || lim(n**3*exp(-n),n,oo) ||.
(4) lim;log(n)**2/n**(1/2);n;oo || lim;(-1)**n;n;oo || lim;(-1)**(2*n);n;oo || lim;1/n**(2/3);n;oo || lim;1/n**(3/2);n;oo ||.

Derivatives are limits by definition. The derivative of a differentiable function \(f\) at \(x\) is defined as \(f'(x)=\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}\) by lim;(f(x+h)-f(x))/h;h;0 ||. Click differentiation to learn more about derivatives.

Examples
(1) The derivative of \(y=x^2\) at \(x\) = 1 is 2, which is defined by \(\lim\limits_{h\to0}\frac{(1+h)^2-1^2}{h}\). Check lim;((1+h)**2-1**2)/h;h;0 ||.
(2) lim;(sin(pi/6+h)-1/2)/h;h;0 || lim;(a**x-1)/x;x;0 || lim;(f(x)-f(5))/(x-5);x;5 ||.

Limit Laws and Properties

Let \(a,b\), and \(c\) be some constants, \(f(x)\) and \(g(x)\) be functions, and \(\displaystyle \lim_{x\to c}f(x)=L, \lim_{x\to c}g(x)=M\). Then
Basic limits \(\displaystyle \lim_{x\to c}a =a, \lim_{x\to c}x=c,\lim_{x\to c}|x|=|c|\).
Constant multiple \(\displaystyle\lim_{x\to c}[af(x)]=a\lim_{x\to c}f(x)=aL\)
Linear combinations \(\displaystyle \lim_{x\to c}[af(x)\pm bg(x)]=a\lim_{x\to c}f(x)\pm b\lim_{x\to c}g(x)=aL\pm bM\)
Product \(\displaystyle\lim_{x\to c}[f(x)g(x)]=[\lim_{x\to c}f(x)][\lim_{x\to c}g(x)]=LM \)
Quotient \(\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\displaystyle\lim_{x\to c}f(x)}{\displaystyle\lim_{x\to c}g(x)}=\frac{L}{M}, M\ne 0.\)
Power \(\displaystyle\lim_{x\to c}[f(x)]^{\frac{p}{q}}=[\lim_{x\to c}f(x)]^{\frac{p}{q}}=L^{\frac{p}{q}}\) where \(p, q\) are integers, \(q\ne 0\), \(L>0\) if \(q\) is even; and \(L\ne 0\) if \(\frac{p}{q} < 0\).
Root \(\displaystyle\lim_{x\to 0}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\to 0}f(x)}=\sqrt[n]{L}\), for any positive integer \(n\) and \(L\ge 0\) if \(n\) is even.

Attention The assumption that the limits of \(f(x)\) and \(g(x)\) must exist matters. Without this assumption, these limit rules do not hold. For example, if \(f(x)=\sqrt{x}, g(x)=x^{\frac{3}{2}}\) and \(f(x)g(x)=x^2\). But the limit \(\displaystyle\lim_{x\to -2}f(x)g(x)=\lim_{x\to -2}x^2=4\) is not correct because both \(f(x)\) and \(g(x)\) are not defined for all \(x\) near -2, and their limits do not exist as \(x\) approaches -2.

Squeeze (sandwich) theorem Let \(f, g, h\) be three functions and \(g(x)\le f(x)\le h(x)\) for all \(x\) in an open interval containing \(c\), but not necessarily at \(x=c\) itself. If \(\displaystyle \lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L\), then \(\displaystyle\lim_{x\to c}f(x)=L\). Similar results hold for limits at infinite and one-sided limits.

Limit inequality Let \(f,g\) be two functions. If \(f(x)\le g(x)\) for all \(x\) in an open interval containing \(c\), except possibly at \(x=c\) itself, then \(\displaystyle\lim_{x\to c}f(x)\le \lim_{x\to c}g(x)\), provided the limit exist.

Limits for composite functions If \(\displaystyle\lim_{x\to c}g(x) = L\) and \(f\) is continuous at \(x=L\), then \(\displaystyle\lim_{x\to c}f[g(x)]=f[\lim_{x\to c}g(x)]\)\(= f(L)\). The result remains valid if \(x\to c\) is replaced by \(x\to c^+, x\to c^-, x\to\infty\) or \(x\to-\infty\).

Evaluating Limits: [lim; f(x); x; c/oo; n/p/np], [lim(f(x), x, c/oo, n/p/np)]

(i) Direction substitution If \(f(x)\) is known to be continuous at \(x = c\), then the limit of \(f(x)\) as \(x \to c\) can be found by substituting \(x = c\) into the formula of \(f(x)\). For example the limit of \(f(x)=2x^2 - 3x\) as \(x \to\) -1 is 5.

(ii) Limit properties are usually used to evaluate the limits of sums, differences, linear combinations, products, quotients, power, roots and composites. For example, the limit of \(f(x)=x^2 - \frac{x}{x^2+1}\) as \(x\) approaches 1 is 1/2.

(iii) Algebraic transformations If substituting \(x = c\) into \(f(x)\) yields an indeterminate form such as \(\frac{0}{0}, \frac{\infty}{\infty}, \infty\cdot 0, \infty-\infty, 1^{\infty}, 0^0 ,\infty^0\), whose limits may or may not exist. But after possible algebra like cancelling common factors or multiplying by conjugates for some expressions, we may transfer \(f(x)\) into a new expression, which may be continuous at \(x = c\).

Examples
(1) Substituting \(x=-2\) into \(\frac{x^2+x-2}{x^2-x-6}\) yields an indeterminate form of \(\frac{0}{0}\). But by factoring both numerator and denominator and canceling the common factor, we have \(\frac{(x-1)(x+2)}{(x-3)(x+2)}=\frac{x-1}{x-3}\), and the limit is \(\frac{3}{5}\), which can be found by substituting \(x=-2\) into the new expression. Verify it by lim;(x**2+x-2)/(x**2-x-6);x;-2 ||.
(2) For fractions involving square roots, first rationalize the numerators or denominators, and then find the limits using some limit properties. Let \(f(x)=\frac{\sqrt{x+9}-3}{x}=\frac{(\sqrt{x+9}-3) (\sqrt{x+9}+3)}{x(\sqrt{x+9}+3)} =\frac{1}{\sqrt{x+9}+3}\). As \(x\to 0,f(x)\to\frac{1}{6}\). Without rationalizing the numerator, we would have an indeterminate form \(\frac{0}{0}\) by direction substitution. Check the limit by lim;((x+9)**(1/2)-3)/x;x;0 ||.

(iv) Variable substitution or change of variables can help find limits at infinity.

Examples
(1) Let \(t=\frac{1}{x}\). Then \(x\to \infty\) is equivalent to \(t\to 0^+\), and \(x\to -\infty\) is equivalent to \(t\to 0^-\). Conversely, \(x\to 0^+ ⇒t\to \infty\) and \(x\to 0^-⇒t\to -\infty\). It follows \(\displaystyle\lim_{x\to0^+}(1+x)^{\frac{1}{x}}= \lim_{t\to \infty}(1+\frac{1}{t})^t=e\), \(\displaystyle\lim_{x\to0^-}(1+x)^{\frac{1}{x}} =\lim_{t\to-\infty}(1+\frac{1}{t})^t=e\). Verify the results by lim;(1+t)**(1/t);t;0;p || lim;(1+t)**(1/t);t;0;n ||.
(2) Let \(t=\frac{1}{x}\). Then \(\displaystyle\lim_{x\to\infty}\frac{2x^2-3x}{x^2+1}=\lim_{t\to 0^+}\frac{2-3t}{1+t^2}=2\), and check it by lim;(2*x**2-3*x)/(x**2+1);x;oo || lim;(2-3*t)/(1+t**2);t;0;p ||.

(v) Limits of rational functions It is important to learn the techniques for finding limits of rational/fraction functions at infinity. Suppose \(f(x)=\frac{P(x)}{Q(x)}=\frac{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+\cdots+b_1x+b_0}\) is a rational function and \(a_n>0\) and \(b_m >0\). Then as \(x\to\infty,P(x)\to \infty,Q(x)\to \infty\), and \(f(x)\) appears to be a form \(\frac{\infty}{\infty}\).

If \(n>m,\displaystyle\lim_{x\to\infty}f(x)=\infty\), and if \(n<m,\displaystyle\lim_{x\to\infty}f(x)=0\).

In particular, if \(n=m\) and \(a_n≠0\) and \(b_m≠0,\displaystyle\lim_{x\to\pm\infty}f(x)=\frac{a_n}{b_m}\). These results are obtained by dividing numerator and denominator the highest power of \(x\) in the fraction.

Examples
(1) As \(x\to\infty,\frac{2x^5-3x}{x^3+4x^2}\to\infty,\frac{15x^2+4}{x^4-13x^2}\to 0,\frac{8x^6-3x^4}{3x^6+7x^5}\to\frac{8}{3}, \frac{4x^7-5x^3}{8x-9x^7}\to-\frac{4}{9}\). Verify th results by lim;(2*x**5-3*x)/(x**3+4*x**2);x;oo || lim;(15*x**2+4)/(x**4-13*x**2);x;oo || lim;(8*x**6-3*x**4)/(3*x**6+7*x**5);x;oo || lim;(4*x**7-5*x**3)/(8*x-9*x**7);x;oo ||.
(2) As \(x\to-\infty,\frac{2x^3-5x}{7x^8+3x^4}\to 0,\frac{2x-x^5}{4x^3-3x^2}\to-\infty,\frac{2x^9-7x^6}{5x^7-6x^5}\to\infty,\frac{3x^3+x^2}{2x^2-1}\to-\infty\). Verify the results by lim;(2*x**3-5*x)/(7*x**8+3*x**4);x;-oo || lim;(2*x-x**5)/(4*x**3-3*x**2);x;-oo || lim;(2*x**9-7*x**6)/(5*x**7-6*x**5);x;-oo || lim;(3*x**3+x**2)/(2*x**2-1);x;-oo ||.

(vi) Limits of fractional functions Suppose \(\frac{P(x)}{Q(x)}\) is not a rational function. Divide numerator \(P(x)\) and denominator \(Q(x)\) the highest power of \(x\).

Examples
(1) \(\displaystyle\lim_{x\to\infty}\frac{2x+1}{\sqrt{x^2+2}}=\lim_{x\to\infty}\frac{2+1/x}{\sqrt{1+2/x^2}}=2\). Check lim;(2*x+1)/(x**2+2)**(1/2);x;oo ||.
(2) \(\displaystyle\lim_{x\to-\infty}\frac{3x+4}{\sqrt{9x^2-5x}}=\lim_{x\to-\infty}\frac{3+4/x}{-\sqrt{9-5/x}}\) = -1. Check lim;(3*x+4)/(9*x**2-5*x)**(1/2);x;-oo ||. Note that in the last equation the numerator is negative and denominator is always positive as \(x\to-\infty\), so the fraction becomes negative. Dividing the denominator (a square root) by a large negative number also means dividing \(-\sqrt{x^2}\).
(3) lim;(x**2-2*x)/(x**4-x)**(1/2);x;oo || lim;(2*x-x**2)/(x**4-x)**(1/2);x;-oo ||.

To find limits of rational/fraction functions at a fixed number, if direct substitution results in an indeterminate form, use algebraic transformation.
(1) Try factoring both numerator and denominator to see if there is any common factor for cancellation.
(2) Try transforming the function by multiplying both numerator and denominator the same conjugate. If both methods fail, use method (3).
(3) Try L'hopital's rule for limits of indeterminate forms.

Examples
(1) \(\displaystyle\lim_{x\to 4}\frac{x-4}{\sqrt{x}-2}=\lim_{x\to 4}\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}-2}\) = 4, lim;(x-4)/(x**(1/2)-2);x;4 ||.
(2) \(\displaystyle\lim_{x\to 1}\frac{\sqrt{x+3}-2}{x^2-x}=\lim_{x\to 1}\frac{(\sqrt{x+3}-2)(\sqrt{x+3}+2)}{x(x-1)(\sqrt{x+3}+2)}\)\(=\lim\limits_{x\to 1}\frac{1}{x(\sqrt{x+3}+2)}=\frac{1}{4}\), lim;((x+3)**(1/2)-2)/(x**2-x);x;1 ||.

(vii) Use import and special limits Use some important limits such as \(\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1,  \lim_{x\to 0}\frac{1-\cos x}{x}=0, \lim_{x\to 0}\frac{\tan x}{x}=1, \lim_{x\to \infty}(1+\frac{1}{x})^x=e\).

Examples Check lim;sin(x)/x;x;0 || lim;(1-cos(x))/x;x;0 || lim;tan(x)/x;x;0 || lim;(1+1/x)**x;x;oo || lim;(1+x)**(1/x);x;0 || lim;sin(2*x)/x;x;0 || lim;tan(3*x)/(4*x);x;0 || lim;(1+2/x)**x;x;oo || lim;(1-3/x)**x;x;oo ||.

Continuity at a point Assume a function \(f\) is defined on an open interval containing \(c\). Then \(f\) is continuous at \(x=c\) if \(\displaystyle \lim_{x\to c}f(x)=f(c)\). The three conditions must be satisfied for the definition:
(1) \(f(c)\) is defined,
(2) the limit \(\displaystyle \lim_{x\to c}f(x)\) exists and is real, and
(3) the limit is equal to \(f(c)\). Otherwise, \(f\) is discontinuous at \(x=c\).

One-sided continuity A function \(f\) is said to be right continuous at \(c\) if \(\displaystyle \lim_{x\to c^+}f(x)=f(c)\), and to be left continuous if \(\displaystyle \lim_{x\to c^-}f(x)=f(c)\). The function \(f\) is continuous at \(c\) if and only if both left and right continuities exist and their limits are equivalent, or \(\displaystyle \lim_{x\to c^+}f(x)=\lim_{x\to c^-}f(x)=f(c)\).

Continuity on an interval If a function \(f\) is continuous at all points in an open interval \(I\), then \(f\) is continuous on \(I\).
If \(f\) is continuous at all values in its domain, \(f\) is a continuous function.

For a function \(f\) defined on a closed interval \(I=[a,b]\), \(f\) is continuous on \([a,b]\) if \(f\) is continuous at each interior point of the interval, is right continuous at \(a\) or \(\displaystyle\lim_{x\to a^+}f(x)=f(a)\), and is left continuous at \(b\) or \(\displaystyle\lim_{x\to b^-}f(x)=f(b)\).

Similarly, \(f\) is continuous on a half open interval \((a, b]\) (or \([a, b)\)) if \(f\) is continuous at each interior point of the interval and is left continuous at \(b\) (or right continuous at \(a\)). If \(f\) is continuous at each interior point of an open interval \((a,b)\), then \(f\) is continuous on \((a,b)\).

Examples Verify continuity at a point or on an interval. Verify continuity of \(f(x) = ⌊x⌋\) on [0, 1] by lim;floor(x);x;0;p || floor(0) ||, which implies \(f\) is right-continuous at 0, and by lim;floor(x);x;1;n || floor(1) ||, which indicates f is not left continuous at 1. In fact, \(f\) is continuous on [0, 1).

Types of discontinuities When a function \(f\) is discontinuous at \(x=c\), neither the limit \(\displaystyle\lim_{x\to c}f(x)\) exists, nor the limit is equal to \(f(c)\).

Removable discontinuity If the limit \(\displaystyle\lim_{x\to c}f(x)=L≠f(c)\) exists but is not equal to \(f(c)\), then we say \(f\) has a removable discontinuity. This implies we can remove the discontinuity by defining \(f\) at \(x=c\) such that \(f(c)=L\). Then the function will be continuous at \(x=c\).

Examples If \(f(x)=\frac{x^2-4}{x-2}\) for \(x\ne 2\), then \(\displaystyle\lim_{x\to 2}f(x)=4\ne f(2)\) because \(f(2)\) is not defined. If we redefine \(f(2)=4\), then \(f\) is continuous at \(x=2\).

In general, if a function \(f\) has a limit \(L\) at point \(c\), where it is not defined, we can define a new function \(F(x)=f(x)\) for \(x\ne c\) and \(F(x)=L\) for \(x=c\). This function \(F\) is the continuous extension of \(f\) at \(x=c\).

Jump discontinuity If both one-sided limits of \(f\) exist as \(x\) approaches \(c\), but they are not equivalent, or \(\displaystyle\lim_{x\to c^+}f(x)\ne \lim_{x\to c^-}f(x)\), we say \(f\) has a jump discontinuity at \(x=c\). For example, the sign function \(\frac{|x|}{x}\) has a jump discontinuity at \(x=0\) because the left-side limit is -1 and the right-side is 1. Note that jump discontinuity is not removable because the limit does not exist.

Infinite discontinuity If one or both one-sided limits of \(f\) is infinite as \(x\) approaches \(c\) , then \(f(x)\) has an infinite discontinuity at \(x=c\), and the line \(x=c\) is called a vertical asymptote to the graph of \(f\).

Examples \(\tan x\) has an infinity discontinuity at \(\frac{π}{2}\) since as \(x →\frac{π}{2}^+,\tan x→\infty\) and as \(x →\frac{π}{2}^-,\tan x→-\infty \). In fact, \(\tan x\) has infinitely many such discontinuities, which occur at \(x=kπ+\frac{π}{2}\) for \(k\) an integer.

Examples There are other types of discontinuities. The Dirichlet function, which is defined by f(x) = 1 if x is rational and f(x) = -1 otherwise, is everywhere discontinuous, or it is not continuous at any point on the entire real line.

Examples The discontinuity of the function \(\sin\frac{1}{x}\) at \(x\) = 0 is not removable, not jump, nor infinite, because its value oscillates between 1 and -1 as x approaches 0.

Properties of continuity If the functions \(f\) and \(g\) are continuous at \(x=c\), then the following functions are also continuous at \(x=c\).
(1) Linear combination \(af\pm bg\) for any constants \(a,b\).
(2) Product \(f\cdot g\).
(3) Quotient \(\frac{f}{g}\) for \(g(c)\ne 0\).
(4) Power \(f^{\frac{p}{q}}\), where \(p,q\) are integers for \(q\ne 0\), and \(f(c)>0\) if \(q\) is even and \(f(c)\ne 0\) is \(q\) is odd.
(5) Root \(f^{\frac{1}{n}}=\sqrt[n]{f}\), where \(n\) is a positive integer and \(f(c)\ge 0\) if \(n\) is even.

These rules also apply to three or more functions. The assumption that both \(f\) and \(g\) are continuous matters. Otherwise, these properties do not hold.

Continuity of composite functions If a function \(g\) is continuous at \(c\) and a function \(f\) is continuous at \(g(c)\), then the composite function \(f\circ g\) is continuous at \(c\). If \(g\) is continuous everywhere and \(f\) is continuous everywhere, then \(f\circ g\) is continuous everywhere. As a result, \(\displaystyle\lim_{x\to c}(f\circ g)(x)=\lim_{x\to c}f[g(x)]=f[\lim_{x\to c}g(x)]=f[g(c)] =(f\circ g)(c)\).

Intermediate value theorem If \(f\) is continuous on a closed interval \([a,b]\) and \(f(a)\ne f(b)\), then for every value \(y\) between \(f(a)\) and \(f(b)\), there exists at least one value \(c\in(a,b)\) such that \(f(c)=y\).

The graphical interpretation for the intermediate theorem is that if \(f\) is continuous on \([a,b]\), then the horizontal line \(y=k\) for \(f(a) < k < f(b)\) intersect the graph of \(f(x)\) at least once.

A fixed point property If \(f\) is continuous on [0, 1] and \(0\le f(x)\le 1\), then there exists at least one point \(c\in[0,1]\) such that \(f(c)=c\), and \(c\) is called a fixed point of \(f\). Let \(g(x)=x-f(x)\). Then this property is a result of the intermediate value theorem for \(g(x)\) on [0, 1].

Extreme value theorem Continuous functions have both minimum and maximum values on closed intervals. If \(f\) is continuous on \([a,b]\), then \(f\) has both minimum \(m\) and maximum \(M\) values on \([a,b]\), or \(m\le f(x)\le M\) for all \(x\in[a,b]\).

Mean value theorem If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists at least one value \(c\in(a,b)\) such that \(f'(c)=\frac{f(b)-f(a)}{b-a}\).

The mean value theorem is fundamental to differential calculus. It relates instantaneous rate of change to average rate of change.

Continuous functions Continuity is a necessary condition for further study of calculus. Elementary functions are continuous on their respective domains. We summarize the continuity of some elementary functions here.
(1) Constants, linear, quadratic, cubic functions and all polynomials are continuous everywhere on the entire real line.
(2) Rational functions \(\frac{p(x)}{q(x)}\) are continuous for all \(x\) such that \(q(x)\ne 0\), where \(p(x)\) and \(q(x)\) are polynomials.
(3) Power/root functions \(f(x)=x^r\) is continuous for all \(x ≥ 0\) and \(r > 0\). Suppose \(r=\frac{m}{n}\) is in simplest form and is rational. If \(n\) is odd, \(f\) is continuous for all \(x≠0\).
(5) Exponential functions \(y=b^x\) are continuous everywhere on the real line for \(b>0, b\ne 1\).
(4) Logarithmic functions \(f(x)=\log_bx\) are continuous for all \(x > 0\) for \(b>0,b\ne 1\).
(6) Sine and cosine are continuous everywhere on the real line. Tangent and secant have infinite discontinuities at \(x=\frac{\pi}{2}+k\pi\) for \(k\) an integer. Cotangent and cosecant have infinite discontinuities at \(x=k\pi\) for \(k\) an integer.
(7) If \(f(x)\) is continuous and one-to-one on its domain, then its inverse \(f^{-1}(x)\) is also continuous on its domain.
(8) If \(g(x)\) is continuous at \(x=c\) and \(f\) is continuous at \(x=g(c)\), then the composite \(f[g(x)]\) is continuous at \(x=c\).

Vertical and horizontal asymptotes The limits of rational functions at infinity depend only on the leading terms of its numerator and denominator. Or \(\displaystyle \lim_{x\to \pm \infty}f(x)=\lim_{x\to \pm \infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1} +\cdots+b_1x+b_0}\)\(=\frac{a_n}{b_m}\lim\limits_{x\to \pm \infty}x^{n-m}\), where \(a_n\ne 0, b_m\ne 0\) and \(n, m\) are nonnegative integers.
If \(n=m, \displaystyle\lim_{x\to \pm \infty}f(x) = \frac{a_n}{b_m}\), and \(y=\frac{a_n}{b_m}\) is a horizontal asymptote.
If \(n < m,\displaystyle\lim_{x\to \pm \infty}f(x) = 0\), and \(y=0\) is a horizontal asymptote.
If \(n=m+1, f(x)\) has an oblique asymptote \(y=ax+b\) and \(\displaystyle \lim_{x\to \pm \infty}[f(x)-(ax+b)]=0\).
If \(n > m\) and \(n-m\) is odd, \(\displaystyle\lim_{x\to \pm \infty}f(x)=\pm \infty\).
If \(n > m\) and \(n-m\) is even, \(\displaystyle\lim_{x\to \pm \infty}f(x)=\infty\). These rules also apply to \(n, m\) that are non-integers.

Asymptotic functions Two curves \(f(x)\) and \(g(x)\) are asymptotic if \(\displaystyle\lim_{x\to\pm\infty}[f(x)-g(x)]=0\).

Examples \(f(x)=\frac{x^2-2}{x-1}=x+1-\frac{1}{x-1}\) and \(g(x)=x+1\) is asymptotic because as \(x\to\pm\infty,f(x)- g(x)\to 0\).

Indeterminate form and L'hopital's rule To find limit by direct substitution, we often obtain the values of a function in form of \(\frac{0}{0}, \frac{\infty}{\infty}, 0\cdot \infty, \infty-\infty, 0^0, 1^{\infty}, \infty^0\), which are called indeterminate forms. If direct substitution results in one of these seven indeterminate forms, the limit may or may not exist, depending on function formulas.

Attention The limit of the forms \(\frac{1}{\infty}, \frac{0}{\infty}, 0^{\infty}\), which are not indeterminate, is equal to 0. These forms \(\infty+1, \infty+\infty\), \(\infty\cdot\infty,\infty^{\infty}, \frac{1}{0^+}\) are all equivalent to \(\infty\). If \(c>0\) is any constant, then \(c\cdot \infty=\infty, c\cdot(-\infty) = -\infty, \frac{\infty}{c}=\infty\),\(\frac{c}{\infty}=0,c-\infty=-\infty\),\(c+\infty=\infty, \infty^c=\infty, c^{\infty}=\infty\) for \(c>1\), and \(c^{\infty}=0\) for \(0< c < 1\).

Examples
(1) If \(f(x)=x,g(x)=x+3\), then as \(x\to \infty, f(x)\to \infty, g(x)\to \infty\), and \(f(x)-g(x)\to 3\). But direct substitution leads to the form \(\infty-\infty\).

(2) As \(x\to \infty\), direct substituting \(x\) in \(h(x)=\sqrt{x+1}-\sqrt{x}\) leads to \(\infty-\infty\). But \(h(x)\to 0\) as \(x\to\infty\). Check it by lim;(x+1)**(1/2)-x**(1/2);x;oo ||.

(3) The limit \(\displaystyle\lim_{x\to\infty}x^3-x\) is \(\infty\) and it has the form \(\infty-\infty\) by direct substitution, and check lim;x**3-x;x; oo ||. Thus, the value of \(\infty-\infty\) can be 3, or 0, or \(\infty\), or its value is indeterminate (not determined).

(4) As \(x\to 0^+, 1+x\to 1, \frac{1}{x}\to\infty\), and \((1+x)^{\frac{1}{x}}\to e\). But direct substitution leads to the form \(1^{\infty}\).

(5) As \(x\to 0, \sin x\to 0, \frac{1}{x}\to\infty\), and \(\frac{\sin x}{x}\to 1\). But direct substitution leads to the form \(0\cdot\infty\) or \(\frac{0}{0}\).

Refer to the mean value theorem to learn more on L'hopital's rule for limits of functions with indeterminate forms.