Question

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

Answer 1

`lim_(xto2)1/(x-2)^2=oo`

Explanation:

Consider the fact that `lim_(xto0^+)1/x=oo`

also notice that

`x=2 => (x-2)^2=(2-2)^2=0^2=0`

So as `xto2` we see that `(x-2)^2 to 0`

so let `u=(x-2)^2`

then

`lim_(uto0^+)1/u=lim_(xto2)1/(x-2)^2`

and since `lim_(xto0^+)1/x=oo`

`lim_(uto0^+)1/u=oo`

and since `u=(x-2)^2` and `uto0` as `xto2`

`lim_(xto2)1/(x-2)^2=oo`