Evaluating limit of exponential function of rational function.

Question

If $$\lim_{x\to 3} \left(\dfrac{\sqrt{2x+3}-x}{\sqrt{x+1}-x+1} \right)^{ \left(\dfrac{x-1-\sqrt{x^2-5}}{x^2-5x+6}\right)}$$ can be expressed in the form $\dfrac{a\sqrt{b}}{c}$, where $a,b,c \in \mathbb{N}$, find the least value of $a^2+b^2+c^2$.

Answer: $29$

I got the limit to be $\dfrac{3\sqrt{10}}{10}$, but that doesn't match with the given answer. Any help will be great. Thanks!

Edit: I'm actually looking for a solution not using L' Hopital.

Answer 1

$$\left(\frac{\sqrt{2x+3}-x}{\sqrt{x+1}-x+1}\right)^{\frac{x-1-\sqrt{x^2-5}}{x^2-5x+6}}=$$ $$=\left(\frac{(2x+3-x^2)(\sqrt{x+1}+x-1)}{(x+1-x^2+2x-1)(\sqrt{2x+3}+x)}\right)^{\frac{(x-1)^2-x^2+5}{(x-2)(x-3)(x-1+\sqrt{x^2-5})}}=$$ $$=\left(\frac{(x+1)(\sqrt{x+1}+x-1)}{x(\sqrt{2x+3}+x)}\right)^{\frac{-2}{(x-2)(x-1+\sqrt{x^2-5})}}\rightarrow\left(\frac{8}{9}\right)^{-\frac{1}{2}}=\frac{3}{2\sqrt2}=\frac{3\sqrt2}{4}.$$

Answer 2

$$\lim_{x\to 3} \left(\frac{\sqrt{2x+3}-x}{\sqrt{x+1}-x+1}\right)^{\frac{x-1-\sqrt{x^2-5}}{x^2-5x+6}} \\ \overset{\text{L.H.}}= \left(\frac 89 \right)^{-\frac 12} \\ =\frac{3\sqrt 2}{4} $$

$$3^2+4^2+2^2 =29$$