# I want to find $\lim\limits_{x\to 5}\frac{2^x-2^5}{x-5}$ without using l'Hopital's rule

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. – Arturo Magidin Jan 20 '11 at 16:06
Why not change "this problem" in the title to your actual problem? It never hurts to be unambiguous in your title. – Jason DeVito Jan 20 '11 at 19:30
@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. – Orbling Jan 20 '11 at 20:51
@Orbling; fair enough. I'll try to be more inviting in the future. – Arturo Magidin Jan 20 '11 at 21:58
@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. – Orbling Jan 20 '11 at 22:03

## 4 Answers

$\lim_{x\to 5}\frac{2^x-2^5}{x-5}=\lim_{x\to 5}\frac{f(x)-f(5)}{x-5}=f'(5)$, 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 $z\to 0$.

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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. – Pete L. Clark Jan 20 '11 at 21:20
@Pete.L.Clark: Well, wait for a moment. I shall fill it up. – anonymous Jan 20 '11 at 21:24
@pete L.Clark: I have added the formula. – anonymous Jan 20 '11 at 21:26
Right, that is the formula. What is the proof? (Hint: see the other answers.) – Pete L. Clark Jan 20 '11 at 21:39
@Pete: I can actually prove it, but i don't think the OP needs that. – anonymous Jan 20 '11 at 21:42