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I am so stuck here. I have no idea where do I start. Could anyone help me out with this:

$$\lim\limits_{h \to 0}\frac{e^{(x+h)\log (a)}-e^{x\log (a)}}{h}$$

copper.hat
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Prakash
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  • This is the derivative of $f(x)=e^{x\log(a)}$ – Christopher King Feb 23 '13 at 15:44
  • Did anybody think that he haven't studied the definition of a derivative yet? – Marra Feb 23 '13 at 15:54
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    @Gustavo: That doesn't mean the OP recognized the limit as a difference quotient, nor that he is used to the idea of using definitions in reverse (i.e. use derivatives to compute the limit, rather than using limits to compute the derivative). (My apologies if I misinterpreted your intent) –  Feb 23 '13 at 15:56
  • Yes, I agree. But I remember, when I first studied calculus, that the authors would sometimes include some exercises using the derivative limit before introducing the definition of derivative. I thought that this could be the case, and it would be misleading since many people are telling him to look at it as a derivative. – Marra Feb 23 '13 at 16:06

6 Answers6

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Can you make a good choice of $f(x)$ ?

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]

cactus314
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Hint: What is $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\quad ?$$Now choose $f(x)$ appropriately and you'll see immediately what to do :)

Clayton
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An idea: Think about the definition of $f'(x)$ by using the limit. and then assume $f(x)=a^x$

Mikasa
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$$\lim\limits_{h \to 0}\frac{e^{(x+h)\log (a)}-e^{x\log (a)}}{h}$$

$$=e^{x\log a}\lim\limits_{h \to 0}\left(\frac{e^{h\log a}-1}h\right)$$

$$=e^{x\log a}\log a\lim\limits_{h \to 0}\left(\frac{e^{h\log a}-1}{h\log a}\right)\text { as } h\to0\implies h\log a\to 0\text{ for finite } a>0 $$

$$=a^x\log a \text { as } e^{m\log a}=(e^{\log_ea})^m=a^m$$

lab bhattacharjee
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  • Don't you still need to show that $$\lim_{u\to 0}\frac{e^u-1}{u}=1$$ – genepeer Feb 23 '13 at 23:00
  • @genepeer, I think following two are too common standard limit formula to demand explanation : http://math.stackexchange.com/questions/152605/proving-that-lim-limits-x-to-0-fracex-1x-1, http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – lab bhattacharjee Feb 24 '13 at 09:29
  • Given that s/he asked a question so many answers deemed as simple, I feel like this was the crucial step to solving it. So my best guess was s/he doesn't know this and it might be helpful to show him/her that. Just a thought, you don't need to explain that since you've given links already. – genepeer Feb 24 '13 at 19:18
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This is the derivative of $a^x$ which equals $a^xln(a)$. Divide both sides by $ln(a)$, take the logarithm on both sides and you're almost done. By the way, it is $\frac{-ln(ln(a))}{ln(a)}$.

  • You assume that the question is asking for a solution in $x$, not in $a$. You assume we want a single value for $x$ which makes the equation true at one point. Not that we want a single value $a$ that makes the equation true at all values $x$. – CogitoErgoCogitoSum Feb 23 '13 at 22:05
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According to your header question, $$\frac{d}{dx} a^x = a^x \ln(a) = 1$$

I dont know if you meant to put that "$=1$" in there, but you did.

There is no solution. There is no singular value for $a$ that makes this equality true for all values of $x$.

CogitoErgoCogitoSum
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    the problem may be ill posed, but you can solve it for x, see my answer above. –  Feb 23 '13 at 19:07
  • I know you can solve $a^x \ln(a)=1$ for $x$. But $x$ is a continuous variable over which the differentiation occurs. I dont see solving for $x$ as valid. To me the question asks for a solution in terms of $a$, which is as I stated not solvable. All you did was solve for a single point $x$ in which the equality holds. I, however, was looking for a value $a$ in which the equality always holds. – CogitoErgoCogitoSum Feb 23 '13 at 22:00
  • Plus, there is no point in me reiterating the work you already gave. Is there? So why are you criticizing the perfectly valid work I gave? – CogitoErgoCogitoSum Feb 23 '13 at 22:05
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    This seems better suited for a comment than an answer. – robjohn Feb 24 '13 at 08:46