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

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Answer 1 · 2017-11-05T07:40:03

The limit is $=lna$

We calculate the limit as follows

$lim_(t->0)(a^t-1)/t=(a^0-1)/0=0/0$

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

$lim_(t->0)(a^t-1)/t=lim_(t->0)((a^t-1)')/(t')$

Let $y=a^t$

Taking logarithm on both sides

$lny=ln(a^t)=tlna$

Differentiating

$dy/y=lna dt$

$dy/dt=ylna=a^tlna$

Therefore,

$lim_(t->0)((a^t-1)')/(t')=lim_(t->0)((a^t '-1')/(t'))$

$=lim_(t->0)((a^tlna-0))/(1)$

$=lna$

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