Science:Math Exam Resources/Courses/MATH110/December 2014/Question 02 (b)/Solution 1

Recall that $\lim _{x\to a}f(x)$ exists if and only if $\lim _{x\to a+}f(x)=\lim _{x\to a-}f(x)$.

First, we evaluate the left hand limit;

Since $x\to 6^{-}$, we consider $x<6$. In this range, $|x-6|=-(x-6)$, so that we get

$\lim _{x\to 6^{-}}{\frac {x^{2}-5x-6}{|x-6|}}=\lim _{x\to 6^{-}}{\frac {(x-6)(x+1)}{-(x-6)}}=\lim _{x\to 6^{-}}{\frac {(x+1)}{-1}}=-7.$

Now, we compute the right hand limit;

Since $x\to 6^{+}$, we consider $x>6$. In this range, $|x-6|=(x-6)$, so that we get

$\lim _{x\to 6^{+}}{\frac {x^{2}-5x-6}{|x-6|}}=\lim _{x\to 6^{+}}{\frac {(x-6)(x+1)}{(x-6)}}=\lim _{x\to 6^{+}}{(x+1)}=7.$

Collecting the terms, we have

$\lim _{x\to 6^{-}}{\frac {x^{2}-5x-6}{|x-6|}}=-7\neq 7=\lim _{x\to 6^{+}}{\frac {x^{2}-5x-6}{|x-6|}}$.

As a result, the limit doesn't exist. $\color {blue}True$.