Science:Math Exam Resources/Courses/MATH110/December 2016/Question 04 (c)/Solution 1

Clearly ${\textstyle f(x)}$ is defined at ${\textstyle x=1}$ and ${\textstyle f(1)=2}$.

The left hand limit is

$${\begin{aligned}\lim \limits _{x\to 1^{-}}f(x)&=\lim \limits _{x\to 1^{-}}{\dfrac {x^{2}-1}{x-1}}\\&=\lim \limits _{x\to 1^{-}}{\dfrac {(x-1)(x+1)}{(x-1)}}\\&=\lim \limits _{x\to 1^{-}}(x+1)\\&=1+1\\&=2.\end{aligned}}$$

Since

$$\lim \limits _{x\to 1^{+}}e^{x}=e^{1}=e$$

and

$$\lim \limits _{x\to 1^{+}}(x-1)=0,$$

the right hand limit is

$${\begin{aligned}\lim \limits _{x\to 1^{+}}f(x)&=\lim \limits _{x\to 1^{+}}{\dfrac {e^{x}}{x-1}}\\&=+\infty .\end{aligned}}$$

This means that ${\textstyle \lim \limits _{x\to 1^{+}}f(x)}$ does not exist. Therefore, ${\textstyle f(x)}$ fails to be continuous at ${\textstyle x=1}$.

Answer: ${\textstyle {\color {blue}f(x){\mbox{ is not continuous at }}x=1{\mbox{ because }}\lim \limits _{x\to 1^{+}}f(x){\mbox{ does not exist.}}}}$