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I got this problem.

Prove that $\lim_{x\to 0}\dfrac{a^x-1}{x}= \ln(a)$ by using the definition of limit and by the fact $\lim_{x\to 0} \dfrac{x}{\log_{a}(1+x)}=\ln(a)$ and without using any other theorem, and of course without using L'Hospital rule.

I got stuck, I don't know which $\delta$ to choose.

Any hints will be appreciated.

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  • $\begingroup$ it is simple if we can useL'Hospital rule. $\endgroup$ – Paul Sep 23 '14 at 6:20
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Note that $$a^x=e^{x\ln a}=\sum_{n=0}^\infty\dfrac{(x\ln a)^n}{n!}.$$ Therefore $$\frac{a^x-1}{x}=\sum_{n=1}^\infty\frac{x^{n-1}(\ln a)^n}{n!}=\ln a+\frac{x(\ln a)^2}{2}+ \dots \to \ln a $$ as $x \to 0.$

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Let us use the substitution $a^x=y+1$ then $x=\log_a(y+1)$ and $y \to 0$ as $x \to 0.$ Then $$\lim_{x\to 0}\dfrac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(y+1)}= \ln(a)$$

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  • $\begingroup$ Thanks, but how can I prove it using the $\epsilon - \delta$ definition of limit and by using the fact that $\lim_{x\to 0} \dfrac{x}{\log_{a}(1+x)}=\ln(a)$ (and without using the limit of sequence/series)? $\endgroup$ – MathNerd Sep 23 '14 at 6:30
  • $\begingroup$ See my edit. You do not wont to use the definition. $\endgroup$ – Bumblebee Sep 23 '14 at 6:40
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Put $x=e^y-1$ and show that $x\rightarrow 0 \implies e^y \rightarrow 1$.

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