Chapter 2.3, Problem 45E

To determine

To calculate: The $\underset{x\to 1}{\lim }f\left(x\right)$ , if $\underset{x\to 1}{\lim }\frac{f\left(x\right)-8}{x-1}=10$

Answer to Problem 45E

  $\underset{x\to 1}{\lim }f\left(x\right)=8$

Explanation of Solution

Given information:

Given $\underset{x\to 1}{\lim }\frac{f\left(x\right)-8}{x-1}=10$

Consider the given limit as,

  $\underset{x\to 1}{\lim }\frac{f\left(x\right)-8}{x-1}=10$

Since, $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)}$

Hence,

  $\begin{array}{c}\frac{\underset{x\to 1}{\lim }\left(f\left(x\right)-8\right)}{\underset{x\to 1}{\lim }\left(x-1\right)}=10\\ \underset{x\to 1}{\lim }\left(f\left(x\right)-8\right)=10\underset{x\to 1}{\lim }\left(x-1\right)\end{array}$

Since, $\underset{x\to a}{\lim }\left(f\left(x\right)\pm g\left(x\right)\right)=\underset{x\to a}{\lim }f\left(x\right)\pm \underset{x\to a}{\lim }g\left(x\right)$

Hence,

  $\begin{array}{c}\underset{x\to 1}{\lim }f\left(x\right)-\underset{x\to 1}{\lim }8=10\left(0\right)\\ \underset{x\to 1}{\lim }f\left(x\right)-8=0\\ \underset{x\to 1}{\lim }f\left(x\right)=8\end{array}$

Therefore, $\underset{x\to 1}{\lim }f\left(x\right)=8$