# `lim_(x->0^+) x^(sqrt(x))` 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...

`lim_(x->0^+) x^(sqrt(x))` 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.

### Textbook Question

Chapter 4, 4.4 - Problem 55 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Let's transform the expression to get a form valid for use of l’Hospital’s Rule.

`x^(sqrt(x)) = (e^(lnx))^(sqrt(x)) = e^(sqrt(x)*lnx) = e^((lnx/(x^(-1/2))))`

Consider the power, `lnx/(x^(-1/2))`

Both numerator and denominator are differentiable for x>0 and both tend to infinity when x tends to zero+. Differentiating them separately, we obtain

`(1/x)/((-1/2)*x^(-3/2)) = -2*sqrt(x),` which tends to zero when x tends to zero+.

Therefore we can apply l’Hospital’s Rule, and the limit of power is 0. Therefore the limit in question is `e^0` = 1.