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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...

Arturo Magidin
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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 04 '12 at 05:24
  • Sorry cost me a lot.. will fix that.. –  Jul 04 '12 at 05:26
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    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 04 '12 at 05:27
  • @ArturoMagidin 2^x +{}^1-8 is what I mean exactly ... that is the excersise ... can you LaTex for me please? –  Jul 04 '12 at 05:28
  • @maniat1k: I can't LaTeX what you write because what you write is $$2^x + {}^1 -8$$ and that is nonsense. The exponent ${}^1$ next to the $+$ is unintelligible. Did you mean $2^{x+1}$? (Note the $+$ is small and in the exponent, not at the same level as the $2$) – Arturo Magidin Jul 04 '12 at 05:30
  • the one that @MarkBennet suggest is the one.. thanks –  Jul 04 '12 at 05:31
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    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 04 '12 at 05:33
  • Do you know about derivatives yet? How do you define exponentials with arbitrary bases? – Arturo Magidin Jul 04 '12 at 05:36
  • do not know about devivates... the exponentials are just like the book says.. –  Jul 04 '12 at 05:38
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    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 04 '12 at 05:42
  • sorry folks it's dificult to me to write down in LateX, and to understand the terms in English ... I'm learning on the go.. I'm speacks spanish ... potencia is exponencial to me, i'm guessing derivatives is derivadas.. but just guessing ... will keep sarching some examples in spanish maybe that is the real solution to me ... thanks anyway –  Jul 04 '12 at 05:47
  • Como define tu libro la funcion exponencial? Y cuales limites haz aprendido a hacer que tienen funciones exponenciales? Definiste $2^x$ usando la exponecial base $e$, o de otra manera? Cual es tu libro de texto? – Arturo Magidin Jul 04 '12 at 05:50
  • 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. –  Jul 04 '12 at 05:58

3 Answers3

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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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Arturo Magidin
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  • Ok entendí... era un cambio de variables... no me daba cuenta de como comenzarlo.. gracias –  Jul 04 '12 at 06:03
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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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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 $ $(a\gt0)\implies -4\lim_{h\to 0}\frac{2^h-1}{h} = -4\ln 2$.

Arturo Magidin
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Aang
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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 04 '12 at 05:44