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**Archie Meade** Alternatively, in terms of you substitution

$\displaystyle \lim_{x\to 0}\left(\frac{a^x-1}{x}\right)$

$\displaystyle a^x-1=t,\ a^x=t+1$

$\displaystyle x=0\ \Rightarrow\ 1=t+1\ \Rightarrow\ t=0$

$\displaystyle lna^x=ln(t+1)$

$\displaystyle xlna=ln(t+1)$

$\displaystyle x=\frac{ln(t+1)}{lna}$

$\displaystyle \frac{1}{x}=\frac{lna}{ln(t+1)}$

$\displaystyle \lim_{x\to 0}\left(\frac{a^x-1}{x}\right)=\left(lna\right)\lim_{t\to 0}\left(\frac{t}{t+1}\right)$

Finally use

$\displaystyle \lim_{t\to 0}\left[ln(t+1)\right]=t$