This is in the section having to do with l'Hospital's Rule and I'm not sure if I should just apply that rule right away because the x^x kind of confused me.
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Let f(x)=(xˣ-1) and g(x)=ln(x)+x-1. xˣ=e^(xln(x)). As x→0⁺, xˣ→1, while ln(x)→-∞, so f(x)/g(x)→0.

[f'(x)=ln(x)xˣ/x and g'(x)=(1/x)+1=(1+x)/x.

L'Hôpital's rule says we can use f'(x)/g'(x)=ln(x)xˣ/(1+x) with the same limit x→0⁺ in f(x)/g(x).

Note, however, that the expression cannot be evaluated for x<0, so the left limit cannot be evaluated. This may invalidate the use of l’Hôpital's rule.

The denominator approaches 1 in this limit. ln(x)→-∞ while xˣ→1. So under this rule the limit would be -∞.]

The graph above appears to show that the right limit is zero. Another way to find the limit is to set x to a small positive value and evaluate the expression. As x gets smaller, while remaining positive, the expression approaches zero.

by Top Rated User (721k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
84,551 questions
89,519 answers
2,000 comments
13,721 users