Chapter 11.2, Problem 58E

(a)

To determine

To approximate: the limitgraphically

Answer to Problem 58E

Here, the value of the limit is $.1$

Explanation of Solution

Given: $\underset{x\to 5^{+}}{\lim }\frac{5-x}{25-x^2}$

Calculation:

Part A In order to approximate this one-sided limit, graph the function and view it within a small widow around $x=5$ . See what value the function approaches as $x$ approaches 5 from the positive side(from the right): see from the graph, the function appears to approach $.1$ as $x\to 5^{+}$ .

Conclusion:

Hence, the limit approaches $.1$

(b)

To determine

To approximate: the limitNumerically

Answer to Problem 58E

Here, the limit approaches $.1$

Explanation of Solution

Calculation:

Now, numerically approximate the limit by using a table of values of $x$ close to 5, and by observing what value the function appears to approach as $x$ decreases to 5 (since this is the one-sided limit $x\to 5^{+}$ ). Follow the directions for your graphing calculator to obtain such a table.

$x$	$f\left(x\right)$
$5.5$   $5.4$   $5.3$   $5.2$   $5.1$   $5.$   $4.9$	$0.0952381$   $0.0961538$   $0.0970874$   $0.0980392$   $0.0990099$ Indeterminate   $0.10101$

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

Conclusion:

Hence,the limit approaches $.1$

(c)

To determine

To evaluate:the limitalgebraically

Answer to Problem 58E

Here, the limit of thefunction is $\frac{1}{10}$ .

Explanation of Solution

Calculation:

Part C In order to evaluate this limit algebraically, first notice that direct substitution of 5 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 $x=5$ is the same as that of the original function, and can be evaluated by direct substitution:

  $\underset{x\to 5^{+}}{\lim }\frac{5-x}{25-x^2}=\underset{x\to 5^{+}}{\lim }\frac{5-x}{\left(5-x\right)\left(5+x\right)}$

  $=\underset{x\to 5^{+}}{\lim }\frac{1}{\left(5+x\right)}$

  $=\frac{1}{10}$

By direct substitution

Therefore, the limit of this function as $x\to 5^{+}$ is $\frac{1}{10}$ .

  

Conclusion:

Hence, the limit of this function is $\frac{1}{10}$ .