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limit as x approaches 0 for the following:

1. (2^x - 1) / x

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

TIA

2006-10-27 11:50:29 · 3 answers · asked by DoughboyFresh 2 in Science & Mathematics Mathematics

3 answers

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

2006-10-27 12:06:39 · answer #1 · answered by Wal C 6 · 1 0

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.

2006-10-27 18:55:43 · answer #2 · answered by MsMath 7 · 0 1

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

2006-10-27 18:53:26 · answer #3 · answered by arbiter007 6 · 0 0

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