# How do you find the limit as x goes to 0 for the function (3^x- 8^x)/ (x)?

Apr 28, 2015

I haven't figured out how to find it algebraically. But here's a non-algebraic solution:

${\lim}_{x \rightarrow 0} \left({3}^{x} - {8}^{x}\right) = 1 - 1 = 0$ and, obviously, ${\lim}_{x \rightarrow 0} x = 0$

So the limit has indeterminate form $\frac{0}{0}$.

Apply l'Hopital's Rule (it is not algebraic, it involves derivatives)

$\frac{d}{\mathrm{dx}} \left({3}^{x} - {8}^{x}\right) = {3}^{x} \ln 3 - {8}^{x} \ln 8$ and, of course $\frac{d}{\mathrm{dx}} \left(x\right) = 1$

So

${\lim}_{x \rightarrow 0} \frac{{3}^{x} - {8}^{x}}{x} = {\lim}_{x \rightarrow 0} \frac{{3}^{x} \ln 3 - {8}^{x} \ln 8}{1} = \ln 3 - \ln 8$

Apr 28, 2015

Remembering the remarkable limit:

${\lim}_{x \rightarrow 0} \frac{{a}^{x} - 1}{x} = \ln a$

than:

${\lim}_{x \rightarrow 0} \frac{{3}^{x} - {8}^{x}}{x} = {\lim}_{x \rightarrow 0} \frac{{3}^{x} - 1 + 1 - {8}^{x}}{x} =$

$= {\lim}_{x \rightarrow 0} \left[\frac{{3}^{x} - 1}{x} - \frac{{8}^{x} - 1}{x}\right] = \ln 3 - \ln 8 = \ln \left(\frac{3}{8}\right)$.