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

Answer 1

Cameron Alexander

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#

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Answer 2

Isaiah Allen

To find the limit of (a^t-1)/t as t approaches 0, we can use L'Hôpital's rule. Taking the derivative of both the numerator and denominator with respect to t, we get (a^t * ln(a))/1. Evaluating this expression as t approaches 0, we have (a^0 * ln(a))/1, which simplifies to ln(a). Therefore, the limit of (a^t-1)/t as t approaches 0 is ln(a).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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