Question

l'hospital's rule help

Can I use l'hospital's rule for the following question

Answer 1

Find the limit for this question \[\lim_{h \rightarrow 0} (4^h-1)/h\]

Answer 2

I do believe you can.

L'Hopital's Rule is valid when you have a function that can be expressed as a quotient, where the limit of the numerator and the denominator functions BOTH evaluate to 0 or BOTH evaluate to infinity

In this case, taking the limit of the numerator as x-> 0 gives you 0. Doing the same for the denominator also gives you 0. So derivate away!

Answer 3

i did, and when i used a calculator to estimate the limit it gave me like 1.3863.

Answer 4

hmm... It's been a while since I've done much calculus.. but I think it should evaluate to 1.
\[\lim_{h \rightarrow 0} \frac{ 4^h - 1 }{ h } = \lim_{h \rightarrow 0} \frac{ (4^h - 1) '}{ h' } =\lim_{h \rightarrow 0} \frac{ 4^h}{ 1 } \]

Answer 5

wait, shouldn't i use the power rule for the top?

Answer 6

You can use l'hospital if it is an Indeterminate form.
l'hospital doesn't work for every thing.

\(\dfrac{d}{dh}4^{h} = 4^{h}\cdot \ln(4)\)

Answer 7

Thank you tk.

Answer 8

yup... its been too long o.o
tk is right