Answer:
[tex]\lim_{x \to\ 1}\frac{1}{(x-1)^2}=\infty[/tex]
vertical asymptote at x=1
Step-by-step explanation:
[tex]\lim_{x \to\ 1}\frac{1}{(x-1)^2}[/tex]
First we find out limit approaches from right
[tex]\lim_{x \to\ 1^+}\frac{1}{(x-1)^2}=\infty[/tex]
BEcause the denominator is a positive quantity because it has square
we find out limit approaches from right
[tex]\lim_{x \to\ 1^-}\frac{1}{(x-1)^2}=\infty[/tex]
so limit is infinity
[tex]\lim_{x \to\ 1}\frac{1}{(x-1)^2}=\infty[/tex]
To find out vertical asymotote , we take the denomintor and set it =0
(x-1)^2=0
Take square on both sides
(x-1) = 0
add 1 on both sides
x=1
So vertical asymptote at x=1
