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#