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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...

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    He utilizado el numero de neper un par de veces pero no las suficientes, 2x es solo letra del ejercicio... mi libro se llama Funciones de Gustavo Duffour.2012-07-04

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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 $ $(a\gt0)\implies -4\lim_{h\to 0}\frac{2^h-1}{h} = -4\ln 2$.

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    Punctuation *inside* displayed equations, *outside* in-line equations. And the OP just said (s)he doesn't know about derivatives, which makes L'Hopital's Rule rather difficult to use.2012-07-04
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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.

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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.

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    Ok entendí... era un cambio de variables... no me daba cuenta de como comenzarlo.. gracias2012-07-04