L'Hospitals Rule?

Question

Lim t-->0 (5^t-3^t)/t the answer is in the book, but how do I get it ln(5/3)?

Answer 1

l'hopital's rule states that you need to differentiate the numerator and denominator separately and then divide them to get the limit

d/dt(5^t - 3^t) = d/dt(t * ln(5) - t * ln(3)) = ln(5) - ln(3) = ln(5/3)

d/dt(t) = 1

therefore, lim t-->0 = ln(5/3)

Answer 2

L'hop is d/dt of top and d/dt of bottem

d(5^t-3^t)=ln(5)*5^t-ln(3)*3^t
d(t)=1
so,
lim t_>0=ln(5)*5^0-ln(3)*3^0=ln(5)-ln(3)=ln(5/3)

Answer 3

Keep in mind d/dt ( a^t ) = a^t * ln(a).

Also, recall that ln(a) - ln(b) = ln(a/b).