I want to find $\lim\limits_{x\to 5}\frac{2^x-2^5}{x-5}$ 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.
 A: First write $2^x$ as $e^{x \ln 2}$. Then let $x=5+t$ so that you get a limit as $t \to 0$. Now you should be able to use the standard limit $(e^z-1)/z \to 1$ as $z\to 0$.
A: $\lim_{x\to 5}\frac{2^x-2^5}{x-5}=\lim_{x\to 5}\frac{f(x)-f(5)}{x-5}=f'(5)$, where $f(x)=2^x$.  $f'(x)=\cdots$ so $f'(5)=\cdots$
A: In principle, you can use L'Hopital's Rule here because both numerator and denominator are differentiable, they both have a limit of $0$ as $x\to 5$, and the limit of the quotient $(2^x-2^5)'/(x-5)'$ exists as $x\to 5$.
Morally, however, you should not use L'Hopital's Rule; the reason is that in order to use L'Hopital's Rule, you need to know what the derivative of $2^x$ is. But the derivative of $2^x$ (at least, at $x=5$) is given precisely by the limit you are trying to do. It would be like using L'Hopital's Rule in order to find $\lim\limits_{\theta\to 0}\frac{\sin\theta}{\theta}$; the problem is that to use L'Hopital's Rule you need ot know the derivative of $\sin\theta$, and in order to know the derivative of $\sin\theta$ you most likely had to figure out this limit in the first place.
So it's a good thing that you don't want to use L'Hopital's Rule here. Instead, you need to recognize this limit as the limit that defines $f'(5)$ with $f(x)=2^x$, and solve it accordingly, whether using the Chain Rule (since $2^x = e^{\ln(2)x}$), or by some other similar method.
A: 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 $$\lim_{x \to 0} \frac{a^{x}-1}{x} = \log{a}$$
