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I want to prove that for $a\gt 0$ and $h\in \mathbb{R}$ the $$\lim_{h \rightarrow 0} \frac{a^h-1}{h}$$ exists. There is no point of definition for a since it is just a real number greater $0$. Since this problem is set in the context of school and the students do not know the $\log$ function, I just wanted to use the properties of the natural exponential function.

I wanted to use a sequence that converges against $0$ and is monotone and bounded. But I'm not sure whether this might work.

The reason to consider this limit in the first place is to determine the derivative of the exponential function $x \mapsto a^x$. However, I am not currently interested in computing this limit or derivative—I just want to show that the limit exists.

Xander Henderson
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Rico1990
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    $\lim\limits_{h\to0} \dfrac{e^h-1}h$ would be $\dfrac d{dx} e^x$ when $x=0$ – J. W. Tanner Apr 21 '20 at 15:43
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    This question lacks enough context to give an appropriate answer. How have you defined $a^h$? What tools can you use? Personally, I think that it would be slightly circular to use L'Hospital's rule (as several of the answers have done), but that is one approach. Another approach is to use properties of the exponential function. What kind of answer are you looking for? Where does this problem come from? Is it from a book? Which book? etc. – Xander Henderson Apr 21 '20 at 17:09
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    Looking at the original version of the question, it appears, perhaps, that you are trying to find the derivative of the function $x \mapsto a^x$, which is computed via the limit you are working with. Is that correct? If so, please edit your question to include that context (in addition to what I requested in my previous comment). – Xander Henderson Apr 21 '20 at 17:11
  • Your question has been closed because it lacks context. If you would like it reopened, please edit it to take into account some (or all) of my critiques. – Xander Henderson Apr 22 '20 at 17:13
  • I edited the question as you asked me to do. – Rico1990 Apr 23 '20 at 12:52
  • I found the answer here: https://math.stackexchange.com/questions/1491220/show-lim-limits-h-to-0-fracah-1h-exists-without-lh%c3%b4pital-or-even-r?rq=1. – Rico1990 Apr 25 '20 at 15:52
  • Thank you for improving your question. I must also commend you for finding a good duplicate target. I spent some time trying to find a good redirect for you a couple of days ago, but most of the posts I found on MSE were about computing this limit using tools which you seemed to want to avoid. Good find. – Xander Henderson Apr 25 '20 at 16:02

2 Answers2

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Set $y=a^x-1$ we get $x=\log_a(y+1)$ then $$ \frac{a^x-1}x=\ln a\left(\frac{y}{\ln(y+1)}\right)=\ln a\left(\frac{1}{\ln(y+1)^{1/y}}\right). $$ now $x\to 0$ if and only if $y\to 0$ and $(1+y)^{1/y}$ is decreasing and locally bounded around $0$ so it has limit when $y$ goes to $0$.

PinkyWay
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Jihlbert
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That depends on exactly how you define $a^x$. If you define it as $\exp(x \ln a)$:

\begin{align*} \lim_{h \to 0} \frac{a^h - 1}{h} &= \lim_{h \to 0} \frac{\exp(h \ln a) - 1}{h} \\ &= \lim_{h \to 0} \frac{\exp(h \ln a) \cdot \ln a}{1} \\ &= \ln a \end{align*}

Here I used l'Hôpital, which by the time you've got to the exponential function and rigurous definition of real powers you should have seen already.

vonbrand
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