trying to understand how to solve this kind of limits. Ok Right now Im doing some excersises... and now I get stuck on this one..
$$\lim_{x\to2}  \frac{2^{x+1}-8}{4-2x}$$
tried to do this 
$$\lim_{x\to2}  \frac{2^{x}\cdot 2-8}{4-2x}$$
but it's the same thing...
 A: After factoring $-2$ from the denominator, and $2$ from the numerator, we have:
$$\lim_{x\to 2}\frac{2^{x+1}-8}{4-2x} = \lim_{x\to 2}\frac{2(2^x - 4)}{-2(x-2)} = -\lim_{x\to 2}\frac{2^x - 4}{x-2}.$$
So we can concentrate on $$\lim\limits_{x\to 2}\frac{2^x-4}{x-2}.$$
We can rewrite this a bit more as
$$\lim_{x\to 2}\frac{2^x - 2^2}{x-2} = 4\lim_{x\to 2}\frac{2^{x-2}-1}{x-2}.$$
Now let $u=x-2$. Then $u\to 0$ as $x\to 2$, so this becomes
$$4\lim_{x\to 2}\frac{2^{x-2}-1}{x-2} = 4\lim_{u\to 0}\frac{2^u-1}{u}.$$
So we are reduced to finding
$$\lim_{u\to 0}\frac{2^u-1}{u}.$$
Rewriting $2^u = e^{\ln(2)u}$ and setting $h = \ln(2)u$, we have that $h\to 0$ as $u\to 0$, so
$$\lim_{u\to 0}\frac{2^u-1}{u} = \lim_{h\to 0}\frac{e^h-1}{h/\ln(2)} = \ln(2)\lim_{h\to 0}\frac{e^h-1}{h}.$$
So we are reduced to finding 
$$\lim_{h\to 0}\frac{e^h-1}{h}.$$
This may be a limit you already know how to do, or else you can find several answers in this site, for example, here. The limit is equal to $1$. Putting it all together, we have:
$$\begin{align*}
\lim_{x\to 2}\frac{2^{x+1}-8}{4-2x} &= -\lim_{x\to 2}\frac{2^x-4}{x-2}\\
&= -4\lim_{x\to 2}\frac{2^{x-2}-1}{x-2}\\
&= -4\lim_{u\to 0}\frac{2^u-1}{u}\\
&= -4\ln(2)\lim_{h\to 0}\frac{e^h-1}{h}\\
&= -4\ln(2).
\end{align*}$$
Although I suspect that you already know how to do some of the limits we found along the way; e.g., if this is an exercise in your book before you know about derivatives, chances are you already know that
$$\lim_{h\to 0}\frac{a^h-1}{h} = \ln(a)$$
when $a\gt 0$. But, since you don't give us enough information, it's impossible to tell.
A: Recall that $$\lim_{x \to 0} \dfrac{\exp(x)-1}{x} = 1$$
The above limit also gives us that $$\lim_{x \to 0} \dfrac{\exp(ax)-1}{x} = a$$
Now lets look at your problem, we have that $$L = \lim_{x \to 2} \dfrac{2^{x+1}-8}{4-2x} = \lim_{x \to 2} \dfrac{2^x-4}{2-x}$$
Replace $x$ by $2+h$. We then get that $$L = \lim_{x \to 2} \dfrac{2^x-4}{2-x} = \lim_{h \to 0} \dfrac{2^{2+h}-4}{-h} = -4 \left(\lim_{h \to 0} \dfrac{2^h-1}{h} \right)$$
Now recall that $2^h = \left(e^{\log 2} \right)^h = e^{h \log 2}$ and make use of the limit at the beginning of this post.
A: Apply L'Hopital's Rule, you will get $$\lim_{x\to 2}\frac{2.2^x.\ln 2}{-2}=-4\ln 2.$$ You can also do it like this: Taking $x=h+2$ gives you $$\lim_{h\to 0}\frac{8(2^h-1)}{-2h}=-4\lim_{h\to 0}\frac{2^h-1}{h}.$$ The limit $\lim_{h\to 0}\frac{a^h-1}{h}=\ln a  $ $(a\gt0)\implies -4\lim_{h\to 0}\frac{2^h-1}{h} = -4\ln 2$.
