We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 5^+}\left(- \sqrt{2} \sqrt{x} + \sqrt{x + 5}\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 5^+}\left(- \sqrt{13 - x} + \sqrt{x + 3}\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 5^+}\left(\frac{- \sqrt{2} \sqrt{x} + \sqrt{x + 5}}{- \sqrt{13 - x} + \sqrt{x + 3}}\right)$$
=
$$\lim_{x \to 5^+}\left(\frac{\frac{d}{d x} \left(- \sqrt{2} \sqrt{x} + \sqrt{x + 5}\right)}{\frac{d}{d x} \left(- \sqrt{13 - x} + \sqrt{x + 3}\right)}\right)$$
=
$$\lim_{x \to 5^+}\left(\frac{\frac{1}{2 \sqrt{x + 5}} - \frac{\sqrt{2}}{2 \sqrt{x}}}{\frac{1}{2 \sqrt{x + 3}} + \frac{1}{2 \sqrt{13 - x}}}\right)$$
=
$$\lim_{x \to 5^+}\left(\frac{\frac{1}{2 \sqrt{x + 5}} - \frac{\sqrt{2}}{2 \sqrt{x}}}{\frac{1}{2 \sqrt{x + 3}} + \frac{1}{2 \sqrt{13 - x}}}\right)$$
=
$$- \frac{\sqrt{5}}{5}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)