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This question already has an answer here:

I'm given a question by my friend. State the value of $\lim_{h\rightarrow 0}\frac{a^h-1}{h}$ for $a=2$.

Then I've used l'Hopital's rule to solve it and I got the answer of $\ln 2$.

But my friend said need to be solved by first principles method. I wonder how. Can anyone show me? Thanks.

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marked as duplicate by Martin Sleziak, SchrodingersCat, Harish Chandra Rajpoot, Davide Giraudo, jameselmore Feb 3 '16 at 14:15

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Hints

  1. Use the definition of the derivative of $f(x)=a^x$ at $x=0$ to construct your limit.

  2. Next, use the differentiation formula $(a^x)'=a^x \ln a$.

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$$\lim_{h\rightarrow 0}\frac{a^h-1}{h}=\lim_{h\rightarrow 0}\frac{e^{\ln a^h}-1}{h}=\lim_{h\rightarrow 0}\frac{e^{h\ln a}-1}{h \ln(a)} \ln(a)\\ =\ln(a)\lim_{s\rightarrow 0}\frac{e^{s}-1}{s}$$

Now identify the last limit as the definition of the derivative of $e^x$ at $x=0$.

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