First for this case we have to check if limit exist or not . So we have to calculate both left hand limit and right hand limit and to exist the limit they has to be same.
LHL:-
$\large \lim_{x->1^{-}}\frac{x}{(x-1)^{2}}$
$\large \lim_{x->1^{-}}$ we are coming toward 1 from from fraction side means say value of x is 0.000000099 like that.
So from the limit we can observe numerator is positive and in denominator (x-1) is negative as value of x is close to 1 from left side . so $\large (x-1)^{2}$ is positive but very small.
so, as x tends to $\large 1^{-}$,then $\large (x-1)^{2}->0^{+}$
So, $\large \lim_{x->1^{-}}\frac{x}{(x-1)^{2}}$ = $\large +\infty$
RHL:-
$\large \lim_{x->1^{+}}\frac{x}{(x-1)^{2}}$
$\large \lim_{x->1^{+}}$ we are coming toward 1 from from right side means say value of x is 1.00000001 like that.
So from the limit we can observe numerator is positive and in denominator (x-1) is positive as value of x is close to 1 from right side . so $\large (x-1)^{2}$ is positive but very small.
so, as x tends to $\large 1^{+}$ ,then $\large (x-1)^{2}->0^{+}$.
$\large \lim_{x->1^{+}}\frac{x}{(x-1)^{2}}$=$\large +\infty$
So,
both left hand limit and right hand limit exists and both are equal to $\large +\infty$ .
So limit exist and the value of the limit is $\large +\infty$.