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

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 $(1)=f'(1)=\sqrt2$.
