Solve limit without L'Hôpital's rule

Question

I am trying to solve the following limit without using L'Hôpital's rule. I know the answer is -8, but I can't seem to figure out another way to solve it.

$$ \lim_{x\to1} \frac{x^2 -1}{2-\sqrt{x+3}} $$

Thanks in advance.

Answer 1

First let $x=u+1$.

$$L=\lim_{u\to0}\frac{(u+1)^2-1}{2-\sqrt{u+4}}=\lim_{u\to0}\frac{u(u+2)}{2-\sqrt{4+u}}$$

By binomial expansion, we have

$$\sqrt{4+u}=2+\frac14u-\frac1{64}u^2+\mathcal O(u^3)$$

Thus, we have

$$L=\lim_{u\to0}\frac{u(u+2)}{2-\sqrt{4+u}}=\lim_{u\to0}\frac{u(u+2)}{-\frac14u+\frac1{64}u^2+\mathcal O(u^3)}=\lim_{u\to0}\frac{u+2}{-\frac14+\frac1{64}u+\mathcal O(u^2)}=\frac2{-\frac14}=-8$$

Answer 2

multiply numerator and denominator by $$2+\sqrt{x+3}$$

Answer 3

HINT:

Set $2-\sqrt{x+3}=u \implies x=(2-u)^2-3,u\to0$

alternatively, $\sqrt{x+3}=v$ $ x=v^2-3$ and $v\to?$

Answer 4

We have $$\frac{x^2 - 1}{2 -\sqrt{x+3}} = \frac{(x^2 - 1)(2 + \sqrt{x+3})}{(2 - \sqrt{x+3})(2+\sqrt{x+3})} = \frac{(x-1)(x+1)(2+\sqrt{x+3})}{1-x}$$

So $$\lim_{x\to 1} \frac{x^2 -1 }{2-\sqrt{x+3}} = -\lim_{x\to 1} (x+1)(2+\sqrt{x+3}) = -8$$