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lim (2^x-1)/(x) x->0
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it is the derivative of \)2^x\) at \(x=0\)
if you got that far
\[\lim_{x \rightarrow 0} (2^x-1)/x\]
L'hopital's rule
oh heck no
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i mean sure you can, but this is the derivative of \(f(x)=2^x\) at \(x=0\) i.e. it is \(f'(0)\)
you know what the derivative of \(y=2^x\) is?
Hint: \[\frac{ d }{ dx } a^x = a^x lnx\]
@iambatman hmmm
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