calculus - I want to find $limlimits_{xto 5}frac{2^x-2^5}{x-5}$ without using l'Hopital's rule

I want to solve this limit without using L'Hopital's rule:
$$\lim_{x\to 5}\frac{2^x-2^5}{x-5}.$$

And thanks.

Answer

You could do this as well. Write $x = 5+h$. So as $x \to 5$, $h \to 0$. Therefore the required limit is

$$\begin{align*} \lim_{h \to 0 } \frac{2^{5+h}-2^{5}}{h} &= 2^{5} \cdot \lim_{h \to 0}\Bigl[ \frac{2^{h}-1}{h}\Bigr] \\ &= 2^{5} \cdot M \end{align*}$$

where $M$= Some value in $\log$. I forgot the formula.

ADDED: The formula is $$\lim_{x \to 0} \frac{a^{x}-1}{x} = \log{a}$$