# 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...

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Sorry, but what is 2^x +^1-8? It can't be $2^x +{}^1-8$, because that does not make sense. Do you mean $x^2 + x - 8$? And what is 2^x*2^1-8? Is it $x^2 +2x - 8$? –  Arturo Magidin Jul 4 '12 at 5:24
Sorry cost me a lot.. will fix that.. –  user35031 Jul 4 '12 at 5:26
I can LaTeX it for you, but I need to understand what you mean. (Is the first limit $$\lim_{x\to 2}\frac{x^2+x-8}{4-2x},$$ or perhaps, as Mark Bennet suggests, $$\lim_{x\to 2}\frac{2^{x+1}-8}{4-2x}$$and the second one $$\lim_{x\to 2}\frac{2^x2-8}{4-2x}\ ?$$ –  Arturo Magidin Jul 4 '12 at 5:27
I deleted my comment - sorry - my computer is not rendering these formulae properly this morning, but Arturo caught it as I put it. –  Mark Bennet Jul 4 '12 at 5:33
I don't know what your book is or what it says. Not everyone uses the same book, you know. What limits of exponential functions have you done, and how? –  Arturo Magidin Jul 4 '12 at 5:42

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.. gracias –  user35031 Jul 4 '12 at 6:03

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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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. –  Arturo Magidin Jul 4 '12 at 5:44