# How do you find the limit of (a^t-1)/t as t->0?

Nov 5, 2017

The limit is $= \ln a$

#### Explanation:

We calculate the limit as follows

${\lim}_{t \to 0} \frac{{a}^{t} - 1}{t} = \frac{{a}^{0} - 1}{0} = \frac{0}{0}$

This is an indeterminate form, so apply l'Hôspital's rule

${\lim}_{t \to 0} \frac{{a}^{t} - 1}{t} = {\lim}_{t \to 0} \frac{\left({a}^{t} - 1\right) '}{t '}$

Let $y = {a}^{t}$

Taking logarithm on both sides

$\ln y = \ln \left({a}^{t}\right) = t \ln a$

Differentiating

$\frac{\mathrm{dy}}{y} = \ln a \mathrm{dt}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = y \ln a = {a}^{t} \ln a$

Therefore,

${\lim}_{t \to 0} \frac{\left({a}^{t} - 1\right) '}{t '} = {\lim}_{t \to 0} \left(\frac{{a}^{t} ' - 1 '}{t '}\right)$

$= {\lim}_{t \to 0} \frac{\left({a}^{t} \ln a - 0\right)}{1}$

$= \ln a$