A calculus problem by Nikola Alfredi

$\begin{aligned} L & = \lim_{n \to 0} \frac {a^n-1}n & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \lim_{n \to 0} \frac {a^n\ln a}1 & \small \blue{\text{Differentiate up and down w.r.t. }n} \\ & = \boxed{\ln a} \end{aligned}$

Reference: L'Hôpital's rule

SOLUTION

So let's look at some other problem and we will try to relate it....

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} e^x = \lim_{\Delta x \rightarrow 0} \ \ [e^x \ \ \frac {e^{\Delta x } - 1}{\Delta x }]$

$\displaystyle \lim_{\Delta x \rightarrow 0} \ \ [e^x \ \ \frac {e^{\Delta x } - 1}{\Delta x }] = e^x \ \ \lim_{\Delta x \rightarrow 0} \frac {e^{\Delta x } - 1}{\Delta x }$

Where, $\ \ \displaystyle \lim_{\Delta x \rightarrow 0} \frac {e^{\Delta x } - 1}{\Delta x } \approx 1.00000.....$

Thus $\ \ \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x$

Now note this $\ \ \displaystyle a = e^{\ln a}$ so $\ \ \displaystyle a^n = e^{n \ln a}$

Hence $\ \ \displaystyle \frac{\mathrm{d}}{\mathrm{d}n} a^n = \frac{\mathrm{d}}{\mathrm{d}n} e^{n \ln a}$

$\ \ \displaystyle a^n \ \ \lim_{\Delta n \rightarrow 0} \frac {a^{\Delta n } - 1}{\Delta n } = e^{n \ln a} \times \ln a \ \$ By Chain-Rule

Ultimately, $\ \ \displaystyle \lim_{\Delta n \rightarrow 0} \frac {a^{\Delta n } - 1}{\Delta n } = \ln a$