Lim8^t-5^t/t as t goes to zero

iv used L'Hospital's Rule because the initial answer is 0/0 ,and then use this fact that e^lnx=x so in this case e^ln8=8 so substituted 8 with that and i got=e^tln8=tln8 as t goes to zero and my final answer is zero but again im not sure about my answer,any help is appreciated :)

You're on the right track.

This can be rewritten as [e^((ln 8)t) - e^((ln 5)t) ] / t

The derivative of the top is (ln 8)e^((ln 8)t) - (ln 5)e^((ln 5)t)

The derivative of the bottom is 1.

At t=0, the derivative of the top becomes (ln 8) - (ln 5), then apply L'Hospital.

lim8^t-5^t/t
By L hospital rule
general formula , d/dt(a^t)= (ln a)a^t

Here a= 8 and 5..
so,
{(ln 8)8^t-(ln 5)5^t}/1
Now apply limit....
=(ln 8)8^0-(ln 5)5^0
=ln 8-ln 5
= 2.079-1.609
=.47

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