# Question #4f6b5

May 15, 2017

Check limit from left and right.

#### Explanation:

Evaluate ${\lim}_{x \to 1} \left(\frac{1 - 2 x}{{x}^{2} - 1}\right)$

Since $x = 1$ is a vertical asymptote (which would mean the limit as $x$ approaches $1$ would be $- \infty$, $\infty$, or does not exist), we need to check the limit from the left and right sides:

Left:
${\lim}_{x \to {1}^{-}} \left(\frac{1 - 2 x}{{x}^{2} - 1}\right)$
When we plug in a number just slightly less than $1$, the limit would have a negative numerator and a negative denominator, therefore the limit as $x$ approaches $1$ from the left is $- \infty$

Right:
${\lim}_{x \to {1}^{+}} \left(\frac{1 - 2 x}{{x}^{2} - 1}\right)$
Wen we plug in a number just slightly greater than $1$, the limit would have a negative numerator and a positive denominator, therefore the limit as $x$ approaches $1$ from the right is $\infty$

Since the limit as $x$ approaches $1$ from the left does not equal the limit as $x$ approaches $1$ from the right, we conclude that the limit does not exist.