Question #4f6b5

1 Answer
May 15, 2017

Check limit from left and right.
Answer: Limit does not exist

Explanation:

Evaluate lim_(x->1)((1-2x)/(x^2-1))

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

Left:
lim_(x->1^-)((1-2x)/(x^2-1))
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 -oo

Right:
lim_(x->1^+)((1-2x)/(x^2-1))
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 oo

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.