MATH 221

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

Last updated: 4 August 2014

Lecture 11

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

Evaluate $\underset{x\to 2}{\text{lim}}\frac{3x^2+8}{x^2-x}\text{.}$

When $x=1,$ $$\frac{3x^2+8}{x^2-x}=11\text{.}$$ When $x=1.5,$ $$\frac{3x^2+8}{x^2-x}=19.66\dots \text{.}$$ When $x=1.9,$ $$\frac{3x^2+8}{x^2-x}=11.011\dots \text{.}$$ When $x=1.99,$ $$\frac{3x^2+8}{x^2-x}=10.091\dots \text{.}$$ When $x=1.999,$ $$\frac{3x^2+8}{x^2-x}=10.00901\dots \text{.}$$ When $x=1.9999,$ $$\frac{3x^2+8}{x^2-x}=10.0009001\dots \text{.}$$ So $$\underset{x\to 2}{\text{lim}}\frac{3x^2+8}{x^2-x}=10\text{.}$$ Usually determining the limit is straightforward.

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

But sometimes...

$\underset{x\to 0}{\text{lim}}\frac{\sqrt{1+x}-1}{x}\stackrel{?}{=}\frac{0}{0}\text{.}$

$\frac{0}{0}$ makes NO SENSE.

$\underset{x\to 0}{\text{lim}}\frac{5x}{x}\stackrel{?}{=}\frac{0}{0}\text{.}$ $$\underset{x\to 0}{\text{lim}}\frac{5x}{x}=\underset{x\to 0}{\text{lim}}5=5\text{.}$$

$\underset{x\to 0}{\text{lim}}\frac{17x}{2x}\stackrel{?}{=}\frac{0}{0}\text{.}$ $$\underset{x\to 0}{\text{lim}}\frac{17x}{2x}=\underset{x\to 0}{\text{lim}}\frac{17}{2}=\frac{17}{2}\text{.}$$

Let's go back to

Evaluate $\underset{x\to 0}{\text{lim}}\frac{\sqrt{1+x}-1}{x}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\sqrt{1+x}-1}{x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{(\sqrt{1+x}-1)}{x}\frac{(\sqrt{1+x}+1)}{(\sqrt{1+x}+1)}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{1+x-1}{x(\sqrt{1+x}+1)}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{x}{x(\sqrt{1+x}+1)}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{1}{\sqrt{1+x}+1}}\\ & =& {\displaystyle \frac{1}{\sqrt{1+0}+1}}\\ & =& {\displaystyle \frac{1}{2}\text{.}}\end{array}$$

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.

$$\underset{x\to 7}{\text{lim}}\frac{x^2-49}{x-7}=\underset{x\to 7}{\text{lim}}\frac{(x-7)(x+7)}{(x-7)}=\underset{x\to 7}{\text{lim}}(x+7)=7+7=14\text{.}$$

Evaluate $\underset{x\to 5}{\text{lim}}\frac{x^5-3125}{x-5}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to 5}{\text{lim}}\frac{x^5-3125}{x-5}}& =& {\displaystyle \underset{x\to 5}{\text{lim}}\frac{x^5-5^5}{x-5}}\\ & =& {\displaystyle \underset{x\to 5}{\text{lim}}\frac{(x-5)(x^4+5x^3+5^2{x}^2+5^{3}x+5^4)}{x-5}}\\ & =& {\displaystyle \underset{x\to 5}{\text{lim}}{x}^4+5x^3+5^2{x}^2+5^{3}x+5^4}\\ & =& {\displaystyle 5^4+5^4+5^4+5^4+5^4=5^5=3125\text{.}}\end{array}$$

Evaluate $\underset{x\to a}{\text{lim}}\frac{x^{\frac{5}{2}}-a^{\frac{5}{2}}}{x-a}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to a}{\text{lim}}\frac{x^{\frac{5}{2}}-a^{\frac{5}{2}}}{x-a}}& =& {\displaystyle \underset{x\to a}{\text{lim}}\frac{(x^{\frac{5}{2}}-a^{\frac{5}{2}})}{(x-a)}\frac{(x^{\frac{5}{2}}+a^{\frac{5}{2}})}{(x^{\frac{5}{2}}+a^{\frac{5}{2}})}}\\ & =& {\displaystyle \underset{x\to a}{\text{lim}}\frac{x^5-a^5}{x-a}·\frac{1}{x^{\frac{5}{2}}+a^{\frac{5}{2}}}}\\ & =& {\displaystyle \underset{x\to a}{\text{lim}}\frac{(x-a)(x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4)}{x-a}\frac{1}{x^{\frac{5}{2}}+a^{\frac{5}{2}}}}\\ & =& {\displaystyle \underset{x\to a}{\text{lim}}\frac{x^4+a{x}^3+a^2{x}^2+a^{3}x+a^4}{x^{\frac{5}{2}}+a^{\frac{5}{2}}}}\\ & =& {\displaystyle \frac{a^4+a^4+a^4+a^4+a^4}{a^{\frac{5}{2}}+a^{\frac{5}{2}}}=\frac{5a^4}{2a^{\frac{5}{2}}}=\frac{5}{2}{a}^{\frac{3}{2}}\text{.}}\end{array}$$

Particularly useful limits

(a) $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}x}{x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^6}{7!}+\frac{x^9}{9!}-\cdots }{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots }\\ & =& {\displaystyle 1-0+0-0+0-\cdots =1\text{.}}\end{array}$$

(b) $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{cos}\hspace{0.17em}x-1}{x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots )-1}{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots }{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}-\frac{x}{2!}+\frac{x^3}{4!}-\frac{x^5}{6!}+\frac{x^7}{8!}-\cdots }\\ & =& {\displaystyle 0+0+0+\cdots =0\text{.}}\end{array}$$

(c) $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{e^x-1}{x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots )-1}{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots }{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\frac{x^4}{5!}+\cdots }\\ & =& {\displaystyle 1+0+0+\cdots =1\text{.}}\end{array}$$

(d) Evaluate $\underset{x\to 0}{\text{lim}}\frac{\text{ln}(1+x)}{x}\text{.}$

Let $y=\text{ln}(1+x)\text{.}$ Then $e^y=1+x$ and $x=e^y-1\text{.}$ Also $y\to 0$ as $x\to 0\text{.}$ So $$\underset{x\to 0}{\text{lim}}\frac{\text{ln}(1+x)}{x}=\underset{y\to 0}{\text{lim}}\frac{y}{e^y-1}=\underset{y\to 0}{\text{lim}}\frac{1}{\frac{e^y-1}{y}}=\frac{1}{1}=1\text{.}$$

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

Evaluate $\underset{n\to \infty }{\text{lim}}{(1+\frac{1}{n})}^n\text{.}$

Let $x=\frac{1}{n}\text{.}$ Then $x\to 0$ as $n\to \infty \text{.}$ So $$\underset{n\to \infty }{\text{lim}}{(1+\frac{1}{n})}^n=\underset{x\to 0}{\text{lim}}{(1+x)}^{\frac{1}{x}}=e\text{.}$$ Note: $n\to \infty$ means $n$ gets larger and larger.

Evaluate $\underset{x\to \pi }{\text{lim}}\frac{\text{sin}\hspace{0.17em}x}{x-\pi }$

Let $y=x-\pi \text{.}$ Then $y\to 0$ as $x\to \pi \text{.}$ So $$\begin{array}{ccc}{\displaystyle \underset{x\to \pi }{\text{lim}}\frac{\text{sin}\hspace{0.17em}x}{x-\pi }}& =& {\displaystyle \underset{y\to 0}{\text{lim}}\frac{\text{sin}(y+\pi )}{y}}\\ & =& {\displaystyle \underset{y\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}y\hspace{0.17em}\text{cos}\hspace{0.17em}\pi +\text{cos}\hspace{0.17em}y\hspace{0.17em}\text{sin}\hspace{0.17em}\pi }{y}}\\ & =& {\displaystyle \underset{y\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}y(-1)+\text{cos}\hspace{0.17em}y·0}{y}}\\ & =& {\displaystyle \underset{y\to 0}{\text{lim}}-\frac{\text{sin}\hspace{0.17em}y}{y}=-1\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 11 from a series of handwritten lecture notes for the class MATH 221 given on October 2, 2000.

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