help with cal 2 - intermediate forms?

Question

limit as x approaches 0 for the following:

1. (2^x - 1) / x

2. [cos x - 1 + (x^2/2)] / x^4

TIA

Answer 1

lim x → 0 (2^x - 1) / x
= lim x → 0 ((ln2 . 2^x )/ 1) (using l'Hopitals rule and 2^x = e^(xln2)
= ln2

lim x → 0 {[cos x - 1 + (x^2/2)] / x^4}
= lim x → 0 {[- sinx + x]/(4x³)} (using l'Hopitals rule)
= lim x → 0 {[- cosx + 1]/(12x²)} (using l'Hopitals rule again)
= lim x → 0 {sin x/(24x)} (using l'Hopitals rule and again)
= 1/24 . lim x → 0 {sin x/(x)}
= 1/24

Answer 2

1.) If you plug in zero you will get
(2^0 - 1)/0 = 0/0 which is an indeterminant form. You can use L'Hopital's rule
Take the derivative of the top: (2^x - 1)' = (ln 2) 2^x
Take the derivative of the bottom: (x)' = 1
Now plug in zero for x
(ln 2) 2^x / 1 = (ln 2)2^0 = ln 2
2) This is similar to #1. When you plug in 0 you get 0/0
Take the derivative of the top and take the derivative of the bottom. Then plug in zero again. If you get an indeterminant form, then repeat.

Answer 3

Looks like you need to apply L'Hospital's rule.