Math 242: Calculus I

1.7 Definition of a Limit

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1.7.1 Definition of Limits

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Definition

The left-hand limit of $f(x)$ as $x$ approaches $a$ is denoted by

$$ \displaystyle\lim_{x\rightarrow a^-} f(x) = L $$

if and only if for every $\epsilon > 0$, there is a $\delta > 0$ so that $a-x<\delta$ implies that $|f(x)-L|<\epsilon$.

The right-hand limit of $f(x)$ as $x$ approaches $a$ is denoted by

$$ \displaystyle\lim_{x\rightarrow a^+} f(x) = M $$

if and only if for every $\epsilon > 0$, there is a $\delta > 0$ so that $x-a<\delta$ implies that $|f(x)-L|<\epsilon$.

Definition

We say that the limit of a function is equal to $L$ as $x$ approaches $a$ if and only if, given an $\epsilon > 0$, we can find a small enough $\delta > 0$ so that $|x-a|<\delta$ implies that $|f(x)-L|<\epsilon$.

Example 1

Consider a constant function $y=2$. Then we claim that $\displaystyle\lim_{x\rightarrow 5^+} = 2$. We need to justify this claim using the Weierstrauss definition of the limit first here and in section 1.5.

Proof

Given a small positive number $\epsilon>0$, we must find a $\delta > 0$ so that $x - a < \delta$ implies that $|f(x)-L| < \epsilon$. In this problem

$$ a = 5,\ L = 2,\ f(x) = 2 $$

We must find a $\delta > 0$ so that $x - 5 < \delta$ implies that $|2 - 2| < \epsilon$. Now $|2 - 2| = 0 < \epsilon$ no matter what $\delta$ we pick. However we must pick one, so let's pick $0 < \delta < \epsilon$. It is preferable to give the reader an interval of $\delta$'s to choose from.

Then for any $x\in\left[5, 5+\delta\right]$, i.e. $x - 5 < \delta$, we have that $|2 - 2| = 0 < \epsilon$.

We just proved that the right-hand limit as $x$ approaches 5 of the constant function $f(x) = 2$ is in fact $2$.

$$ \displaystyle\lim_{x\rightarrow 5^+} 2 = 2 $$

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Exercise 1

Show that the limit from the left of the function $f(x)=2$ as $x$ approaches $5$ is also $2$.

Solution

The answer is very similar. Given an $\epsilon > 0$, if $0 < \delta < \epsilon$, then $5 - x < \delta$ implies that $|f(x) - L| = |2 - 2| = 0 < \epsilon$. Hence $$ \displaystyle\lim_{x\rightarrow 5^-} 2 = 2 $$

In light of Theorem 1.6.2, and the fact that the right-hand and left-hand limits have the same value, we conclude that the two-sided limit exists and that the limit is also $2$.

$$ \displaystyle\lim_{x\rightarrow 5} 2 = 2 $$

Exercise 2

Using the Weierstrauss definition of limit, show that the limit of $f(x)=2$ as $x$ approaches $5$ is $2$.

Solution

The answer is very similar. Given an $\epsilon > 0$, if $0 < \delta < \epsilon$, then $|x-5| < \delta$ implies that $|f(x) - L| = |2 - 2| = 0 < \epsilon$. Hence $$ \displaystyle\lim_{x\rightarrow 5} 2 = 2 $$

Limits of Constant Functions

I hope that everyone see that the value of $5$ is not special. For a constant function and any real number $a$,

$$ \displaystyle\lim_{x\rightarrow a} 2 = 2 $$

Furthermore, the number $2$ is not special.

Theorem 1.7.1

For any constant function $f(x) = k$, where $k\in\mathbb{R}$, and any real number $a\in\mathbb{R}$,

$$ \displaystyle\lim_{x\rightarrow a} k = k $$

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1.7.2 The Identity Function

Example 2

Show that the limit as $x$ approaches $3$ of the function $f(x)=x$ is $3$.

Solution

Given an $\epsilon > 0$, we need a $\delta > 0$ so that $|x-a|<\delta$ implies that $|f(x)-L|<\epsilon$. In this problem

$$ a = 3,\ L = 3,\ f(x)=x $$

This makes

$$ \begin{align*} |x-a| &= |x-3|,\ \text{and} \\ |f(x)-L| &= |x - 3| \end{align*} $$

This tells us that we can pick any number $0<\delta<\epsilon$ just like we did in the previous section! Choose $0 < \delta < \epsilon$, then $|x-3|<\delta$ implies that $|x-3|<\delta<\epsilon$.

In light of Theorem 1.6.2 we also have the left-hand and right-hand limits.

$$ \begin{align*} \displaystyle\lim_{x\rightarrow 3^+} x &= 3 \\ \\ \displaystyle\lim_{x\rightarrow 3^-} x &= 3 \end{align*} $$

As in the previous section the number $3$ in the domain is not special.

Theorem 1.7.2

If $a\in\mathbb{R}$ is any real number, then

$$ \displaystyle\lim_{x\rightarrow a^-} = \displaystyle\lim_{x\rightarrow a^+} = \displaystyle\lim_{x\rightarrow a} = a $$

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1.7.3 Finding Limits

Using Theorems 1.6.1 - 1.6.5, we can now compute many, many limits.

Exercise 3

Compute the limit as $u$ approaches $3$ of the function $r(u) = \frac{1}{u}$.

Solution

$$ \displaystyle\lim_{u\rightarrow 3}\,\frac{1}{u} = \frac{\displaystyle\lim_{u\rightarrow 3}\,1}{\displaystyle\lim_{u\rightarrow 3}\,u} = \frac{1}{3} $$

Exercise 4

Show using the Weierstrauss definition limit that $\displaystyle\lim_{x\rightarrow 3} \frac{1}{u} = \frac{1}{3}$.

Solution

Given an $\epsilon > 0$, we must find a $\delta > 0$ so that $|u-3|<\delta$ implies that $$ \left|\frac{1}{u} - \frac{1}{3}\right| < \epsilon $$ In order to find our interval for $\delta$, we need to start with the result and work backwards toward our inequality $|u-3|<\delta$. $$ \begin{align*} \left|\frac{1}{u} - \frac{1}{3}\right| &< \epsilon \\ \\ \left|\frac{3}{3u} - \frac{u}{3u}\right| &< \epsilon \\ \\ \left|\frac{3 - u}{u}\right| &< 3\epsilon \\ \\ \frac{|u-3|}{|u|} &< 3\epsilon \end{align*} $$ Now this tricky. Remember that we want $|u-3|<\delta$, a really small number. This means that $$ \begin{align*} -\delta &< u-3 < \delta \\ 3-\delta &< u < 3+\delta \\ \end{align*} $$ Remember also that we want $\delta$ to be small, so how about $\delta < \epsilon$. Then $$ 3-\epsilon < 3-\delta < u < 3+\delta < 3+\epsilon $$ So we can replace the demoninator with something smaller, and the quotient will be bigger. $$ \begin{align*} \frac{|u-3|}{|u|} &< \frac{|u-3|}{3-\epsilon} < 3\epsilon \\ \\ |u-3| &< 3\epsilon(3-\epsilon) \end{align*} $$ Now we created two consitions on $\delta$: $$ \begin{align*} \delta &< \epsilon \\ \\ \delta &< 3\epsilon(3-\epsilon) \end{align*} $$ For algebra steps to work in reverse, we need both conditions to be met. So we write:
Choose $0 < \delta < \min\left\{ \epsilon,\ 3\epsilon(3-\epsilon) \right\}$. Then $|u-3|<\delta$ implies that $$ \begin{align*} |u-3| &< 3\epsilon(3-\epsilon) \\ \\ \frac{|u-3|}{|u|} &< \frac{|u-3|}{3-\epsilon} < 3\epsilon \\ \\ \frac{|u-3|}{3|u|} &< \epsilon \\ \\ \left| \frac{u-3}{3u} \right| &< \epsilon \\ \\ \left| \frac{u}{3u} - \frac{3}{3u} \right| &< \epsilon \\ \\ \left| \frac{1}{3} - \frac{1}{u} \right| &< \epsilon \\ \\ \left| \frac{1}{u} - \frac{1}{3} \right| &< \epsilon \\ \end{align*} $$ We just proved that $$ \displaystyle\lim_{u\rightarrow 3}\,\frac{1}{u} = \frac{1}{3} $$

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1.7.4 Infinite Limits

Definition

The limit of a function as $x$ approaches $a$ is $+\inf$ if and only if for every large number $M>0$, there is a $\delta>0$ so that $|x-a|<\delta$ implies that $f(x) > M$.

The limit of a function as $x$ approaches $a$ is $-\inf$ if and only if for every large number $M>0$, there is a $\delta>0$ so that $|x-a|<\delta$ impies that $-f(x) > M$.

Example 3

Using the Weierstrauss definition of an infinite limit, show that the one-sided limit as $x$ approaches $0$ from the right of the function $f(x) = \frac{1}{x}$ is $+\infty$.

Solutions

Given a BIG number $M = 100$, we must find a $\delta > 0$ so that whenever $x - a < \delta$, we have that $f(x) > M$. Here

$$ a = 0,\ M=100,\ f(x) = \frac{1}{x} $$

We need to find a $\delta > 0$ small enough so that $x - 0 < \delta$ implies that $\frac{1}{x} > 100$. As before we start with the conclusion we need and work backwards

$$ \begin{align*} \frac{1}{x} &> 100 \\ \\ x &< \frac{1}{100} \\ \\ x - 0 &< \frac{1}{100} \end{align*} $$

We have our $\delta$! If we chooose $0 < \delta < \frac{1}{100}$, then $x - 0 < \delta$ implies that

$$ \begin{align*} x &< \frac{1}{100} \\ \\ \frac{1}{x} &> 100 \end{align*} $$

done!

Exercise 5

Use the Weierstrauss definition of an infinite limit to show that $\displaystyle\lim_{x\rightarrow 0}\frac{1}{x}$ does not exist.

Solution

To show that a two-sided limit does not exist, we need show that the limit from the right and the limit from the left are different. We already showed in Example 3, that $$ \displaystyle\lim_{x\rightarrow 0^+}\frac{1}{x} = +\infty $$
Let us find the left-hand limit. Given a real number $M > 0$, we need a $\delta > 0$ so that $0 - x < \delta$ implies that $f(x) < -M$. We already have such a $\delta$ from example 3. If we choose $0 < \delta < \frac{1}{M}$. Then $0 - x < \frac{1}{M}$ implies that
$$ \begin{align*} -x &< \frac{1}{M} \\ \\ x &> -\frac{1}{M} \\ \\ \frac{1}{x} &< -M \end{align*} $$ Since we can be given any $M$, no matter how big, we have that as $x$ approaches $0$ from the left, the function $\frac{1}{x}$ gets larger than $M$. Thus
$$ \displaystyle\lim_{x\rightarrow 0^-} \frac{1}{x} = -\infty $$
Since the right-hand limit is $+\infty$ and the left-hand limit is $-\infty$, we have that the two-side limit does not exist.

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1.7.5 Application

Exercise 6

A machinist is required to manufacture a circular disk with area $1000\ \text{cm}^2$.

(a) What radius produces such a disk?

(b) If the machinist is allowed an error tolerance of $\pm\,5\ \text{cm}^2$ in the area of the disk, how close to the ideal radius from part (a)must the machinist control the radius?

(c) In terms of the $\epsilon$,$\delta$ definition of $\displaystyle\lim_{x\rightarrow a}f(x)=L$, what is $x$? What is $f(x)$? What is $a$? What is $L$? What value of $\epsilon$ is given? What is the corresponding $\delta$?

Solution

(a) Area $= 1000\ \text{cm}^2 = \pi\,r^2$, where $r$ is the radius. So $$ r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{1000}{\pi}} = 10\sqrt{\frac{10}{\pi}} \approx 17.84124 $$
(b) In this problem area $A$ is a function of the radius of the disk, $A = A(r) = \pi\,r^2$. A tolerance of $\pm\,5\ \text{cm}^2$ in area means that if our ideal area $L = 1000\ \text{cm}^2$ is the goal, and $r$ is the radius that the machinist and the lathe can make using the available materials, then $-5 < A(r) - L < 5\ \text{cm}^2$.
$$ \begin{align*} |A(r) - L| &< 5 \\ \\ \left| \pi\,r^2 - 1000 \right| &< 5 \\ \\ -5 &< \pi\,r^2 - 1000 < 5 \\ \\ 995 &< \pi\,r^2 < 1005 \\ \\ \frac{995}{\pi} &< r^2 < \frac{1005}{\pi} \\ \\ \sqrt{\frac{995}{\pi}} &< r < \sqrt{\frac{1005}{\pi}} \\ \\ 17.80 &< r < 17.89 \\ \\ 17.80 &< 17.84 < 17.89 \\ \\ -.04 &< \delta < .05 \\ \\ |\delta| &< .04 \end{align*} $$
We should check our algebra. $$ \begin{align*} A(17.80) &= \pi\,17.80^2 \approx 995.38 \\ \\ A(17.88) &= \pi\,17.88^2 \approx 1004.35 \end{align*} $$
These computations tell us that if the machinist keeps the radius within $.04\ \text{cm}$ of $17.84\ \text{cm}$, or $.4\ \text{mm}$ the ideal radius of approximately $17.84\ \text{cm}$, then the area of the disk will be between $995.38\ \text{cm}^2$ and $1004.35\ \text{cm}^2$. This means it will be with $\pm 5\ \text{cm}^2$ of the ideal area of $1000\ \text{cm}^2$.
(c) $$ \begin{align*} x &= r \\ f(x) &= A(r) = \pi\,r^2 \\ a &\approx 17.84\ \text{cm} \\ L &= 1000\ \text{cm}^2 \\ \epsilon &= 5\ \text{cm}^2 \\ \delta &= .4\ \text{mm} \end{align*} $$

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