Chapter 11.2, Problem 57E

(a)

To determine

To approximate: the limit by graphing utility.

Answer to Problem 57E

The function appears to approach $.5$ as $x\to 1^{-}$ .

Explanation of Solution

Given:

  $\underset{x\to -1^{-}}{\lim }=\frac{x-1}{x^2-1}$

Calculation:

In order to approximate this one-sided limit, graph the function and view it within a small window around $x=1$ . See what value the function approaches as

  $x$ approaches $1$ from the negative side (from the left):

  

From the graph, the function appears to approach $.5$ as $x\to 1^{-}$ .

Conclusion:

Therefore, the function appears to approach

  $.5$ as $x\to 1^{-}$ .

(b)

To determine

To approximate: the limit by creating the table.

Answer to Problem 57E

$x$	$f\left(x\right)$
$0.5$	$0.666667$
$0.6$	$0.625$
$0.7$	$0.588235$
$0.8$	$0.555556$
$0.9$	$0.526316$
$1.$	Indeterminate
$1.1$	$0.47619$

Explanation of Solution

Calculation:

Now, try to numerically approximate the limit by using a table of values of $x$ close to $1$ , and by observing what value the function appears to approach as $x$ increases to $1$ (since this is the one-sided limit $x\to 1^{-}$ ). Follow the directions for your graphing calculator to obtain such a table.

$x$	$f\left(x\right)$
$0.5$	$0.666667$
$0.6$	$0.625$
$0.7$	$0.588235$
$0.8$	$0.555556$
$0.9$	$0.526316$
$1.$	Indeterminate
$1.1$	$0.47619$

The table also shows that the function appears to approach the value $.5$ as $x\to 1^{-}$ .

Conclusion:

Therefore, the table is created to approximate the limit.

(c)

To determine

To evaluate: the limit algebraically.

Answer to Problem 57E

The limit of this function is $\frac{1}{2}$ .

Explanation of Solution

Calculation:

In order to evaluate this limit algebraically, first notice that direct substitution of $1$ into the function as written results in the indeterminate form: $\frac{0}{0}$

Instead, evaluate the limit by factoring the denominator, and dividing out common factors in the numerator and denominator in order to provide a function whose limit at $x=1$ is the same as that of the original function, and can be evaluated by direct substitution:

  $\begin{array}{l}\underset{x\to 1^{-}}{\lim }\frac{x-1}{x^2-1}=\underset{x\to 1^{-}}{\lim }\frac{x-1}{\left(x-1\right)\left(x+1\right)}\\ =\underset{x\to 1^{-}}{\lim }\frac{1}{\left(x+1\right)}\\ =\frac{1}{2} Bydirectsubstitution\end{array}$

Before, the limit of this function as $x\to 1^{-}$ is $\frac{1}{2}$ .

  

Conclusion:

Therefore, the limit of this function is $\frac{1}{2}$ .