Exercise on limit of indeterminate form: $\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt2$

Question

I'm not able to show that

$\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt{2}$

I proceeded as follows:

Substituting 1 to $x$ gives the indeterminate form $\frac{0}{0}$.

I rationalized by multiplying the numerator and the denominator by $\sqrt{1+x} + \sqrt{2}$ but I was not able to prosecute, since I obtained again an indetermiante form $\frac{0}{0}$.

I used

$y = x -1$

$x = y + 1$

and computed the limit for $y \rightarrow 0$ and then I rationalized the denominator, but I was again stuck with the indeterminate form $\frac{0}{0}$.

The exercise is taken from a high school book and the chapter comes before the one about de l'Hopital rule, so it must be solved without using it.

Answer 1

If you rationalize both the numerator and the denominator you will end up with $\frac {2(\sqrt {1+x}+\sqrt 2)} {\sqrt {x+3}+\sqrt {5-x}}$ so the limit is $\frac {2(\sqrt2 +\sqrt 2)} {\sqrt 4 +\sqrt 4}=\sqrt {2}$.

Answer 2

If you rationalize the denominator of your expression, the limit becomes$$\lim_{x\to1}\frac{\left(\sqrt{x+3}-\sqrt{5-x}\right)\left(\sqrt{1+x}+\sqrt2\right)}{x-1}.\tag1$$Now define $f(x)=\left(\sqrt{x+3}-\sqrt{5-x}\right)\left(\sqrt{1+x}+\sqrt2\right)$, and then $(1)=f'(1)=\sqrt2$.