Limits

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 27 October 2009

Limits

Write $$\underset{x\to 2}{\lim }f\left(x\right)=10$$ if $f\left(x\right)$ gets closer and closer to $10$ as $x$ gets closer and closer to $2$.

Example. Evaluate ${\displaystyle \underset{x\to 2}{\lim }\frac{3x^2+8}{x^2-x}}$.
When $x=1.5$, ${\displaystyle \frac{3x^2+8}{x^2-x}=19.66\mathrm{\cdots }}$.
When $x=1.9$, ${\displaystyle \frac{3x^2+8}{x^2-x}=11.011\mathrm{\cdots }}$.
When $x=1.99$, ${\displaystyle \frac{3x^2+8}{x^2-x}=10.091\mathrm{\cdots }}$.
When $x=1.999$, ${\displaystyle \frac{3x^2+8}{x^2-x}=10.00901\mathrm{\cdots }}$.
When $x=1.9999$, ${\displaystyle \frac{3x^2+8}{x^2-x}=10.0009001\mathrm{\cdots }}$.

Usually determining the limit is straightforward.

Example. $\underset{x\to 1}{\lim }6x^2-4x+3=5$.

But sometimes …

Example. ${\displaystyle \underset{x\to 1}{\lim }\frac{\sqrt{1+x}-1}{x}\stackrel{?}{=}\frac{0}{0}}$.
Hoewver ${\displaystyle \frac{0}{0}}$ MAKES NO SENSE.

Example. ${\displaystyle \underset{x\to 0}{\lim }\frac{5x}{x}\stackrel{?}{=}\frac{0}{0}}$.
However here we have ${\displaystyle \underset{x\to 0}{\lim }\frac{5x}{x}=\underset{x\to 0}{\lim }5=5}$.

Example. ${\displaystyle \underset{x\to 0}{\lim }\frac{17x}{2x}\stackrel{?}{=}\frac{0}{0}}$.
However here we have ${\displaystyle \underset{x\to 0}{\lim }\frac{17x}{2x}=\underset{x\to 0}{\lim }\frac{17}{2}=\frac{17}{2}}$.

Example. ${\displaystyle \underset{x\to 1}{\lim }\frac{\sqrt{1+x}-1}{x}\stackrel{?}{=}\frac{0}{0}}$.
However here we have $$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-1}{x}& =& \underset{x\to 0}{\lim }\frac{\sqrt{1+x}-1}{x}·\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}\\ & =& \underset{x\to 0}{\lim }\frac{1+x-1}{x\left(\sqrt{1+x}+1\right)}\\ & =& \underset{x\to 0}{\lim }\frac{x}{x\left(\sqrt{1+x}+1\right)}\\ & =& \underset{x\to 0}{\lim }\frac{1}{\sqrt{1+x}+1}\\ & =& \frac{1}{\sqrt{1+0}+1}\\ & =& \frac{1}{2}\text{.}\end{array}$$ So whenever a limit looks like it is coming out to $\frac{0}{0}$ it needs to be looked at in a different way to see what it is really getting closer and closer to.

Example. Evaluate ${\displaystyle \underset{x\to 7}{\lim }\frac{x^2-49}{x-7}}$.
Here we have $$\underset{x\to 7}{\lim }\frac{x^2-49}{x-7}=\underset{x\to 7}{\lim }\frac{\left(x-7\right)\left(x+7\right)}{x-7}=\underset{x\to 7}{\lim }x+7=7+7=14\text{.}$$

Example. Evaluate ${\displaystyle \underset{x\to 5}{\lim }\frac{x^5-3125}{x-5}}$.
Here we have $$\begin{array}{rcl}\underset{x\to 5}{\lim }\frac{x^5-3125}{x-5}& =& \underset{x\to 5}{\lim }\frac{x^5-5^5}{x-5}\\ & =& \underset{x\to 5}{\lim }\frac{\left(x-5\right)\left(x^4+5x^3+5^2{x}^2+5^{3}x+5^4\right)}{x-5}\\ & =& \underset{x\to 5}{\lim }\left(x^4+5x^3+5^2{x}^2+5^{3}x+5^4\right)\\ & =& 5^4+5^4+5^4+5^4+5^4\\ & =& 5^5\\ & =& 3125\text{.}\end{array}$$

Example. Evaluate ${\displaystyle \underset{x\to 5}{\lim }\frac{x^{5/2}-a^{5/2}}{x-a}}$.
Here we have $$\begin{array}{rcl}\underset{x\to 5}{\lim }\frac{x^{5/2}-a^{5/2}}{x-a}& =& \underset{x\to 5}{\lim }\frac{x^{5/2}-a^{5/2}}{x-a}·\frac{x^{5/2}+a^{5/2}}{x^{5/2}+a^{5/2}}\\ & =& \underset{x\to 5}{\lim }\frac{x^5-a^5}{x-a}·\frac{1}{x^{5/2}+a^{5/2}}\\ & =& \underset{x\to 5}{\lim }\frac{\left(x-a\right)\left(x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4\right)}{x-a}·\frac{1}{x^{5/2}+a^{5/2}}\\ & =& \underset{x\to 5}{\lim }\frac{x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4}{x^{5/2}+a^{5/2}}\\ & =& \frac{a^4+a^4+a^4+a^4+a^4}{a^{5/2}+a^{5/2}}\\ & =& \frac{5a^4}{2a^{5/2}}\\ & =& \frac{5}{2}{a}^{3/2}\text{.}\end{array}$$

Particularly useful limits

Example. Evaluate ${\displaystyle \underset{x\to 0}{\lim }\frac{\sin x}{x}}$.
Here we have $$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{\sin x}{x}& =& \underset{x\to 0}{\lim }\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots }{x}\\ & =& \underset{x\to 0}{\lim }\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots \right)\\ & =& 1-0+0-0+0-\cdots \\ & =& 1\end{array}$$

Example. Evaluate ${\displaystyle \underset{x\to 0}{\lim }\frac{\cos x-1}{x}}$.
Here we have $$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{\cos x-1}{x}& =& \underset{x\to 0}{\lim }\frac{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots \right)-1}{x}\\ & =& \underset{x\to 0}{\lim }\frac{-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots }{x}\\ & =& \underset{x\to 0}{\lim }\left(-\frac{x}{2!}+\frac{x^3}{4!}-\frac{x^5}{6!}+\frac{x^7}{8!}-\cdots \right)\\ & =& -0+0-0+0-\cdots \\ & =& 0\text{.}\end{array}$$

Example. Evaluate ${\displaystyle \underset{x\to 0}{\lim }\frac{e^x-1}{x}}$.
Here we have $$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{e^x-1}{x}& =& \underset{x\to 0}{\lim }\frac{\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots \right)-1}{x}\\ & =& \underset{x\to 0}{\lim }\frac{x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots }{x}\\ & =& \underset{x\to 0}{\lim }\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\cdots \right)\\ & =& 1+0+0+0+\cdots \\ & =& 1\text{.}\end{array}$$

Example. Evaluate ${\displaystyle \underset{x\to 0}{\lim }\frac{\ln \left(1+x\right)}{x}}$.
Let $y=\ln \left(1+x\right)$. Then $$e^y=1+x\Rightarrow x=e^y-1\text{,}\text{and}y\to 0\text{ as }x\to 0\text{.}$$ So $$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{\ln \left(1+x\right)}{x}& =& \underset{y\to 0}{\lim }\frac{y}{e^y-1}\\ & =& \underset{y\to 0}{\lim }\frac{1}{\left(\frac{e^y-1}{y}\right)}\\ & =& \frac{1}{1}\\ & =& 1\text{.}\end{array}$$

Example. Evaluate ${\displaystyle \underset{x\to 0}{\lim }{\left(1+x\right)}^{1/x}}$.
We have $$\begin{array}{rcl}\underset{x\to 0}{\lim }{\left(1+x\right)}^{1/x}& =& \underset{x\to 0}{\lim }{\left(e^{\ln \left(1+x\right)}\right)}^{1/x}\\ & =& \underset{x\to 0}{\lim }{e}^{\frac{1}{x}\ln \left(1+x\right)}\\ & =& \underset{x\to 0}{\lim }{e}^{\frac{\ln \left(1+x\right)}{x}}\\ & =& e^1\\ & =& e\text{.}\end{array}$$

Note. $n\to \infty$ means as $n$ gets larger and larger.

Example. Evaluate ${\displaystyle \underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n}$.
Let $x=\frac{1}{n}$. Then $x\to 0$ as $n\to \infty$. So $$\underset{\mathrm{n}\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=\underset{x\to 0}{\lim }{\left(1+x\right)}^{1/x}=e\text{.}$$.

Example. Evaluate ${\displaystyle \underset{x\to \pi }{\lim }\frac{\sin x}{x-\pi }}$.
Let $y=x-\pi$. Then $y\to 0$ as $x\to \pi$. So $$\begin{array}{rcl}\underset{x\to \pi }{\lim }\frac{\sin x}{x-\pi }& =& \underset{y\to 0}{\lim }\frac{\sin \left(y+\pi \right)}{y}\\ & =& \underset{y\to 0}{\lim }\frac{\sin y\cos \pi +\cos y\sin \pi }{y}\\ & =& \underset{y\to 0}{\lim }\frac{\sin \left(y\right)\left(-1\right)+\cos \left(y\right)0}{y}\\ & =& \underset{y\to 0}{\lim }\frac{-\sin \left(y\right)}{y}\\ & =& 1\end{array}$$

Example. Evaluate ${\displaystyle \underset{x\to \infty }{\lim }\frac{x^2-7x+11}{3x^2+10}}$.
We have $$\begin{array}{rcl}\underset{x\to \infty }{\lim }\frac{x^2-7x+11}{3x^2+10}& =& \underset{x\to \infty }{\lim }\frac{1-\frac{7}{x}+\frac{11}{x^2}}{3+\frac{10}{x^2}}\\ & =& \frac{1-0+0}{3+0}\\ & =& \frac{1}{3}\text{.}\end{array}$$

Example. Evaluate ${\displaystyle \underset{x\to 0}{\lim }\frac{\sin 3\mathrm{x}}{\sin 5x}}$.
We have $$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{\sin 3\mathrm{x}}{\sin 5x}& =& \underset{x\to 0}{\lim }\frac{\sin 3x}{3x}·3x·\frac{5x}{\sin 5x}·\frac{1}{5x}\\ & =& \underset{x\to 0}{\lim }\frac{\sin 3x}{3x}·\frac{1}{\left(\frac{\sin 5x}{5x}\right)}·\frac{3x}{5x}\\ & =& 1·\frac{1}{1}·\frac{3}{5}\\ & =& \frac{3}{5}\text{.}\end{array}$$

Example. Evaluate ${\displaystyle \underset{x\to 1}{\lim }\frac{1-x}{{\left(\arccos x\right)}^2}}$.
Let $y=\arccos x$. Then $y\to 0$ as $x\to 1$ and $x=\cos y$. So $$\begin{array}{rcl}\underset{x\to 1}{\lim }\frac{1-x}{{\left(\arccos x\right)}^2}& =& \underset{y\to 0}{\lim }\frac{1-\cos y}{y^2}\\ & =& \underset{y\to 0}{\lim }\frac{1-\cos y}{y^2}·\frac{1+\cos y}{1+\cos y}\\ & =& \underset{y\to 0}{\lim }\frac{1-{\cos }^{2}y}{y^2}·\frac{1}{1+\cos y}\\ & =& \underset{y\to 0}{\lim }\frac{\sin y}{y}·\frac{\sin y}{y}·\frac{1}{1+\cos y}\\ & =& 1·1·\frac{1}{2}\\ & =& \frac{1}{2}\text{.}\end{array}$$

Example. Evaluate ${\displaystyle \underset{\Delta x\to 0}{\lim }\frac{f\left(x+\Delta x\right)-f\left(x\right)}{\Delta x}}$ when $f\left(x\right)=\sin 2x$.
We have $$\begin{array}{rcl}\underset{\Delta x\to 0}{\lim }\frac{f\left(x+\Delta x\right)-f\left(x\right)}{\Delta x}& =& \underset{\Delta x\to 0}{\lim }\frac{\sin \left(2\left(x+\Delta x\right)\right)-\sin \left(2x\right)}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\frac{\sin \left(2x+2\Delta x\right)-\sin \left(2x\right)}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\frac{\sin 2x\cos 2\Delta x+\cos 2x\sin 2\Delta x-\sin 2x}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\sin 2x·\frac{\cos 2\Delta x-1}{\Delta x}+\cos 2x·\frac{\sin 2\Delta x}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\sin 2x·\frac{\cos 2\Delta x-1}{2\Delta x}·2+\cos 2x·\frac{\sin 2\Delta x}{2\Delta x}·2\\ & =& \sin 2x·0·2+\cos 2x·1·2\\ & =& 2\cos 2x\text{.}\end{array}$$

Example. Evaluate ${\displaystyle \underset{\Delta x\to 0}{\lim }\frac{f\left(x+\Delta x\right)-f\left(x\right)}{\Delta x}}$ when $f\left(x\right)=\cos x^2$.
We have $$\begin{array}{rcl}\underset{\Delta x\to 0}{\lim }\frac{f\left(x+\Delta x\right)-f\left(x\right)}{\Delta x}& =& \underset{\Delta x\to 0}{\lim }\frac{\cos \left({\left(x+\Delta x\right)}^2\right)-\cos x^2}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\frac{\cos \left(x^2+2x\Delta x+{\left(\Delta x\right)}^2\right)-\cos x^2}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\frac{\cos x^2\cos \left(2x\Delta x+{\left(\Delta x\right)}^2\right)-\sin x^2\sin \left(2x\Delta x+{\left(\Delta x\right)}^2\right)-\cos x^2}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\cos x^2·\frac{\cos \left(2x\Delta x+{\left(\Delta x\right)}^2\right)-1}{\Delta x}-\sin x^2·\frac{\sin \left(2x\Delta x+{\left(\Delta x\right)}^2\right)}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\cos x^2·\frac{\cos \left(2x\Delta x+{\left(\Delta x\right)}^2\right)-1}{2x\Delta x+{\left(\Delta x\right)}^2}·\frac{2x\Delta x+{\left(\Delta x\right)}^2}{\Delta x}\\ & & -\sin x^2·\frac{\sin \left(2x\Delta x+{\left(\Delta x\right)}^2\right)}{2x\Delta x+{\left(\Delta x\right)}^2}·\frac{2x\Delta x+{\left(\Delta x\right)}^2}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\cos x^2·\frac{\cos \left(2x\Delta x+{\left(\Delta x\right)}^2\right)-1}{2x\Delta x+{\left(\Delta x\right)}^2}·\left(2x+\Delta x\right)\\ & & -\sin x^2·\frac{\sin \left(2x\Delta x+{\left(\Delta x\right)}^2\right)}{2x\Delta x+{\left(\Delta x\right)}^2}·\left(2x+\Delta x\right)\\ & =& \cos x^2·0·2x-\sin x^2·1·2x\\ & =& -2x\sin x^2\text{,}\end{array}$$ (since $2x\Delta x+{\left(\Delta x\right)}^2\to 0$ as $\Delta x\to 0$).

Example. Evaluate ${\displaystyle \underset{\Delta x\to 0}{\lim }\frac{f\left(x+\Delta x\right)-f\left(x\right)}{\Delta x}}$ when $f\left(x\right)=x^x$.
We have $$\begin{array}{rcl}\underset{\Delta x\to 0}{\lim }\frac{f\left(x+\Delta x\right)-f\left(x\right)}{\Delta x}& =& \underset{\Delta x\to 0}{\lim }\frac{{\left(x+\Delta x\right)}^{x+\Delta x}+x^x}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\frac{{\left(e^{\ln \left(x+\Delta x\right)}\right)}^{x+\Delta x}+{\left(e^{\ln \left(x\right)}\right)}^x}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\frac{e^{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)}+e^{x\ln \left(x\right)}}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }{e}^{x\ln x}·\frac{e^{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}+1}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }{e}^{x\ln x}·\frac{e^{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}+1}{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}·\frac{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }{e}^{x\ln x}·\frac{e^{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}+1}{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}\left(\frac{x\ln \left(x+\Delta x\right)-x\ln x}{\Delta x}+\ln \left(x+\Delta x\right)\right)\\ & =& \underset{\Delta x\to 0}{\lim }{e}^{x\ln x}·\frac{e^{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}+1}{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}\left(\frac{x\ln \left(x\left(1+\frac{\Delta x}{x}\right)\right)-x\ln x}{\Delta x}+\ln \left(x+\Delta x\right)\right)\\ & =& \underset{\Delta x\to 0}{\lim }{e}^{x\ln x}·\frac{e^{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}+1}{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}\left(\frac{x\left(\ln x+\ln \left(1+\frac{\Delta x}{x}\right)\right)-x\ln x}{\Delta x}+\ln \left(x+\Delta x\right)\right)\\ & =& \underset{\Delta x\to 0}{\lim }{e}^{x\ln x}·\frac{e^{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}+1}{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}\left(\frac{x\ln \left(1+\frac{\Delta x}{x}\right)}{\Delta x}+\ln \left(x+\Delta x\right)\right)\\ & =& \underset{\Delta x\to 0}{\lim }{e}^{x\ln x}·\frac{e^{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}+1}{\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x}\left(\frac{\ln \left(1+\frac{\Delta x}{x}\right)}{\frac{\Delta x}{x}}+\ln \left(x+\Delta x\right)\right)\\ & =& e^{x\ln x}·1·\left(1+\ln x\right)\\ & =& x^x+x^x\ln x\text{.}\end{array}$$ (since $\left(x+\Delta x\right)\ln \left(x+\Delta x\right)-x\ln x\to 0$ and $\Delta x/x\to 0$ as $\Delta x\to 0$).

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)