Calculate limit $\lim_{x \to 2 \pm 0} \frac{(x+1)^2}{2-x}$ without L'Hopital's rule

Question

I have $\lim\limits_{x \to 2 \pm 0} \frac{(x+1)^2}{2-x}$

I calculated $\lim\limits_{x \to 2 + 0} \frac{(x+1)^2}{2-x}=\lim\limits_{x \to 2 - 0} \frac{(x+1)^2}{2-x}=\lim\limits_{x \to 2 } \frac{(x+1)^2}{2-x}=\frac90=\infty$

Or is it wrong?

Answer 1

No, this isn't correct. The left- and right-side limits are not equal.

$\lim \limits_{x\to 2^-}\frac{(x+1)^2}{2-x} = \left[\frac{9}{0^+}\right] = +\infty$

$\lim \limits_{x\to 2^+}\frac{(x+1)^2}{2-x} = \left[\frac{9}{0^-}\right] = -\infty$

Answer 2

For a different look at it, set $x-2=t$ to get $$ - \lim_{t \to 0} \frac{(t+3)^2}{t} = -6 - \lim_{t \to 0} \frac{9}{t} $$ Clearly as $t \to 0$, this limit does not exist.