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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.

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    (i) Please use mark-up, not ASCII art. (ii) Please ask a question, not just express your desires. (iii) Please do not rely on the title as an integral part of your message: the body should be self-contained. (iv) Please say what you have done or why you seem to be having trouble.2011-01-20
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    Why not change "this problem" in the title to your actual problem? It never hurts to be unambiguous in your title.2011-01-20
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    @Arturo Magidin: (ii)/(iii)/(iv) I agree with you. However, on (i), not everyone knows the syntax, and I think it should be fine to type in standard code-like algebra if you do not know it; with the assumption someone that does will come and represent it properly with an edit.2011-01-20
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    @Orbling: I mention (i) for two reasons: many know it, they just don't know that it is available here. As someone who has been in `sci.math` for well over 15 years, I know all about using ASCII art, and many may not realize it is available here. And those who don't know it ask what that means, and there are links in the editing window that help with it. So it's a prompt to get people to "join in" using it.2011-01-20
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    @Arturo Magidin: Those reasons are sound and good, it just came over (to me) as a rebuke, rather than a suggestion, even with the "please" prefixes.2011-01-20
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    @Orbling; fair enough. I'll try to be more inviting in the future.2011-01-20
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    @Arturo Magidin: Apologies if I've come over as attacking. Just I tend to worry about new users getting a lot of rebukes or corrections on their first arrival. Very common problem across all of SE.2011-01-20
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    Isn't it L'Ho__s__pital's rule?2011-02-09
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    @bobobobo: At least in the U.S., it's commonly spelled both as L'Hospital and L'Hôpital (and hence L'Hopital when people are unable to type an ô). See [the MathWorld article on L'Hôpital's Rule](http://mathworld.wolfram.com/LHospitalsRule.html), about the 2nd or 3rd paragraph, for a bit more info.2011-05-06

4 Answers 4

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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 $$\lim_{x \to 0} \frac{a^{x}-1}{x} = \log{a}$$

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    -1: the evaluation of the limit that you "forgot" is the entire content of the problem.2011-01-20
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    @Pete.L.Clark: Well, wait for a moment. I shall fill it up.2011-01-20
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    @pete L.Clark: I have added the formula.2011-01-20
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    Right, that is the formula. What is the proof? (Hint: see the other answers.)2011-01-20
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    @Pete: I can actually prove it, but i don't think the OP needs that.2011-01-20
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$\lim_{x\to 5}\frac{2^x-2^5}{x-5}=\lim_{x\to 5}\frac{f(x)-f(5)}{x-5}=f'(5)$, where $f(x)=2^x$. $f'(x)=\cdots$ so $f'(5)=\cdots$

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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.

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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 $z\to 0$.