`lim_(x->0^+) (x^x - 1)/(ln(x) + x - 1)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule...

`lim_(x->0^+) (x^x - 1)/(ln(x) + x - 1)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

Asked on by enotes

Textbook Question

Chapter 4, 4.4 - Problem 36 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

1 Answer | Add Yours

shumbm's profile pic

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

The denominator has limit `-oo+0-1=-oo.`

The numerator has limit 0 (I'll prove this below).

Therefore there is no indeterminacy, `0/oo=0.`

(zero is the answer).

Now prove that `x^x->1` when `x->0^+:`

`x^x = (e^lnx)^x = e^(x*lnx).`

`x*lnx = lnx/(1/x),` limit of this exists by l'Hospital's Rule: `oo/oo,` everything is differentiable for x>0, `f'/g' = (1/x)/(-1/x^2) = -x -> 0.`

We’ve answered 318,917 questions. We can answer yours, too.

Ask a question