Chapter B, Problem 1E

To determine

To find: $\underset{x\to 5}{\lim }\left(\frac{x^2-25}{2x-10}\right)$ .

Answer to Problem 1E

  $\underset{x\to 5}{\lim }\left(\frac{2x-0}{2-0}\right)=5$

Explanation of Solution

Given information:

  $\underset{x\to 5}{\lim }\left(\frac{x^2-25}{2x-10}\right)$

Concept:

According to L’hopital’s rule,

If $\underset{x\to c}{\lim }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)$ , reduces to $\frac{0}{0}$ or $\frac{\infty }{\infty }$

Also, if $f\left(x\right)$ and $g\left(x\right)$ are differentiable,

  $\underset{x\to c}{\lim }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\underset{x\to c}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

Calculation:

Given $\underset{x\to 5}{\lim }\left(\frac{x^2-25}{2x-10}\right)$

As the given limit is in the form of $\frac{0}{0}$ . Apply L’hopital’s Rule.

Now differentiate function in numerator and denominator.

  $\underset{x\to 5}{\lim }\left(\frac{x^2-25}{2x-10}\right)=\underset{x\to 5}{\lim }\left(\frac{2x-0}{2-0}\right)$

Now put $x=5$

  $\begin{array}{c}\underset{x\to 5}{\lim }\left(\frac{2x-0}{2-0}\right)=\frac{2\left(5\right)}{2}\\ =5\end{array}$

Conclusion:

  $\underset{x\to 5}{\lim }\left(\frac{2x-0}{2-0}\right)=5$