How do you find the limit of  1 / (x-2)^2 as x approaches 2?

Jan 20, 2017

${\lim}_{x \to 2} \frac{1}{x - 2} ^ 2 = \infty$

Explanation:

Consider the fact that ${\lim}_{x \to {0}^{+}} \frac{1}{x} = \infty$

also notice that

$x = 2 \implies {\left(x - 2\right)}^{2} = {\left(2 - 2\right)}^{2} = {0}^{2} = 0$

So as $x \to 2$ we see that ${\left(x - 2\right)}^{2} \to 0$

so let $u = {\left(x - 2\right)}^{2}$

then

${\lim}_{u \to {0}^{+}} \frac{1}{u} = {\lim}_{x \to 2} \frac{1}{x - 2} ^ 2$

and since ${\lim}_{x \to {0}^{+}} \frac{1}{x} = \infty$

${\lim}_{u \to {0}^{+}} \frac{1}{u} = \infty$

and since $u = {\left(x - 2\right)}^{2}$ and $u \to 0$ as $x \to 2$

${\lim}_{x \to 2} \frac{1}{x - 2} ^ 2 = \infty$