When applying the definition of a derivative to $\frac{d}{dx}b^x$ and a little algebra one arrives to

$$b^x\times\lim\limits_{h \to 0}\frac{b^h - 1}{h}$$

where of course that limit equals $\ln(b)$. I know this limit can be evaluated with L'Hospitals rule, but that involves using the derivative what I am just about to prove, so that would be a circular proof. I suspect one has to reduce this limit to something that relates to the definition of e as $$\lim\limits_{n \to \infty}(1+1/n)^n$$ but I can not do it.

How to show that that limit equals $\ln(b)$, so the proof of the derivative of $b^x$ is complete ?

Note: I found that by substituting for $s = b^h-1$, etc, one can separate $\ln(b)$, but than the another indeterminate limit remains: $\lim\limits_{s \to 0}(s/\ln(1+s))$, which must be $1$, and again I do not want to solve it using L'Hospital's rule, but I cannot otherwise.

  • $\begingroup$ Definitely a duplicate question. $\endgroup$ – Graphth Oct 21 '12 at 15:44
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    $\begingroup$ Both the limit with $h \to 0$ and $s \to 0$ are "elementary" limits which you can derive, for example, with a Taylor expansion of the function. For the limit with $s$ try to bound it from above with something that approaches 1: what do you know about $\log(1 + s)$ in $(0,1)$? $\endgroup$ – Andy Oct 21 '12 at 15:45

As I said in the comment, your substitution is good. Now if we take $\log(s +1)$ can we bound it by something resembling $s$? Recall that $s+1 < e^s$ for all $s > 0$ (try to prove it). Then you can use the fact that if $\alpha \leq \lim_{x \to c} f(x) \leq \lim_{x \to c} g(x) = \alpha$ and the limit of $f$ exists, then $\lim_{x \to c} f(x) = \alpha$ (which is a special case of the comparison theorem).


As I recall we wrote $b^x=e^{x \ln b}$ and used the chain rule: $\frac {d\ b^x}{dx}=\frac {d\ e^{x \ln b}}{dx}=e^{x \ln b}\ln b=\ln b \ b^x$. Have you proved the chain rule yet?


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