# Limits and Their Properties-1.2
[TOC]
## 1.2 Finding Limits Graphically and Numerically
### An Introduction to Limits
To sketch the graph of the function

$$f(x)= \frac{x^3-1}{x-1}$$

for values other than $x = 1$, you can use standard curve-sketching techniques. At $x = 1$, however, it is not clear what to expect.

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To get an idea of the behavior of the graph of f near $x = 1$, you can use two sets of x-values–one set that approaches 1 from the left and one set that approaches 1 from the right, as shown in the table.

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The graph of f is a parabola that has a hole at the point (1, 3), as shown.

$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$

Although x cannot equal 1, you can move arbitrarily close to 1, and as a result f(x) moves arbitrarily close to 3.

Using limit notation, you can write $$\lim_{x\to 1}f(x)=3$$
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This is read as “the limit of f(x) as x approaches 1 is 3.”
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This discussion leads to an informal definition of limit.
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If $f(x)$ becomes arbitrarily close to a single number $L$ as $x$ approaches $c$ from either side, the limit of $f(x)$ as $x$ approaches $c$ is $L$.

This limit is written as $$\lim_{x\to c}f(x)=L$$

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**Example 1 – Estimating a Limit Numerically**
Evaluate the function $$f(x) = \frac{x}{\sqrt{x+1}-1} $$ at several x-values near 0 and use the results to estimate the limit.
$$\lim_{x\to 0}\frac{x}{\sqrt{x+1}-1}.$$
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:::spoiler ***solution.***
The table lists the values of f(x) for several x-values near 0.


From the results shown in the table, you can estimate the limit to be 2.

This limit is reinforced by the graph of f as shown.

$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$
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### Limits That Fail to Exist

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**Example 3 – Different Right and Left Behavior**
Show that the limit $$\lim_{x\to 0}\frac{|x|}{x}$$ does not exist.

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:::spoiler ***solution.***
Consider the graph of the function
$$f(x) = \frac{|x|}{x}$$

In Figure and from the definition of absolute value,

$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$
you can see that

\begin{equation}
\frac{|x|}{x} = \left\{
\begin{array}{lc}
1, \quad x>0 \\
-1, \quad x<0
\end{array} \right.
\end{equation}

So, no matter how close x gets to 0, there will be both positive and negative x-values that yield $f(x) = 1$ or $f(x) = –1$.

Specifically, if $\delta$ (the lowercase Greek letter delta) is a positive number, then for x-values satisfying the inequality $0 < |x| < \delta$ , you can classify the values of $\frac{|x|}{x}$ as –1 or 1 on the intervals.

$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$

Because $\frac{|x|}{x}$ approaches a different number from the right side of 0 than it approaches from the left side, the limit does not exist.
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### A Formal Definition of Limit
Let’s take another look at the informal definition of limit. If $f(x)$ becomes arbitrarily close to a single number $L$ as $x$ approaches $c$ from either side, then the limit of $f(x)$ as $x$ approaches $c$ is $L$, is written as

At first glance, this definition looks fairly technical. Even so, it is informal because exact meanings have not yet been given to the two phrases
$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$ “$f(x)$ becomes arbitrarily close to $L$”
and
$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$ “$x$ approaches $c$.”.

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The first person to assign mathematically rigorous meanings to these two phrases was Augustin-Louis Cauchy. His $\epsilon$-$\delta$ definition of limit is the standard used today.

In the Figure , let $\epsilon$ (the lower case Greek letter epsilon) represent a (small) positive number.

$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$

Then the phrase “$f(x)$ becomes arbitrarily close to $L$” means that $f(x)$ lies in the interval $(L –\epsilon, L +\epsilon)$.

Using absolute value, you can write this as
$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$$|f(x)-L|<\epsilon$

Similarly, the phrase “x approaches c” means that there exists a positive number $\delta$ such that x lies in either the interval $(c-\delta,c)$ or the interval $(c,c+\delta)$

This fact can be concisely expressed by the double inequality
$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$$0<|x|<\delta$

The first inequality $0<|x-c|$ expresses the fact that $x\ne c$.

The second inequality $|x-c|<\delta$ says that x is within a distance $\delta$ of c.


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**Example 6 – Finding a $\delta$ for a Given $\epsilon$**
Given the limit $$\lim_{x\to 3}(2x-5)=3$$ find $\delta$ such that $|(2x-5)-1|<0.01$ whenever $0<|x-3|<\delta$.

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:::spoiler ***solution.***
In this problem, you are working with a given value of $\epsilon$ –namely, $\epsilon = 0.01$.

To find an appropriate $\delta$ , try to establish a connection between the absolute values $|(2x-5)-1|$ and $|x-3|$.

Notice that $|(2x-5)-1|=|2x-6|=2|(x-3)|$

Because the inequality $|(2x-5)-1|<0.01$ is equivalent to $2|x-3|<0.01$,
you can choose $\delta = \frac{1}{2}(0.01) = 0.005$.

This choice works because $0<|x-3|<0.005$
implies that $|(2x-5)-1|=2|x-3|<2(0.005)=0.01$

As you can see in Figure , for x-values within $0.005$ of $3 (x ≠ 3)$, the values of $f (x)$ are within $0.01$ of $1$.

$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$
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###### tags: `Calculus (I)`