PLEASE HELP URGENT!!Use graphs and tables to find the limit and identify any vertical asymptotes of the function. limit of 1 divided by the quantity x minus 1 squared as x approaches 1.

Answer:[tex]\lim_{x \to\ 1}\frac{1}{(x-1)^2}=\infty[/tex]vertical asymptote at x=1Step-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 squarewe 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=0Take square on both sides(x-1) = 0add 1 on both sidesx=1So vertical asymptote at x=1