I tried:

$$\lim\_{x \rightarrow 5} \frac{e^{x-5}-1}{x-\sqrt{4x+5}} = \frac{e^{x-5}-1}{x-\sqrt{4x+5}} \cdot \frac{x+\sqrt{4x+5}}{x+\sqrt{4x+5}} = \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{x^2-(4x+5)} = \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{x^2-4x-5} = \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{x(x-4-\frac{5}{x})} =\\ \frac{(e^{x-5}-1)}{x} \cdot \frac{(x+\sqrt{4x+5})}{(x-4-\frac{5}{x})} $$

if I make $y = x-5$ I get 

$$ \frac{(e^{y}-1)}{y+5} \cdot \frac{(x+\sqrt{4x+5})}{(x-4-\frac{5}{x})} = \frac{(e^{y}-1)}{y} \cdot \frac{1}{1+\frac{5}{y}} \cdot \frac{(x+\sqrt{4x+5})}{(x-4-\frac{5}{x})} = \frac{1}{1+\frac{5}{y}} \cdot \frac{(x+\sqrt{4x+5})}{(x-4-\frac{5}{x})} = ???$$

What do I do next? Am I solving it correctly so far?