💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# 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.$\displaystyle \lim_{x\to 0^+} x^{\sqrt{x}}$

## $$\begin{array}{l}y=x^{\sqrt{x}} \Rightarrow \ln y=\sqrt{x} \ln x, \text { so } \\\lim _{x \rightarrow 0^{+}} \ln y=\lim _{x \rightarrow 0^{+}} \sqrt{x} \ln x=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-1 / 2}} \stackrel{\Perp}{=} \lim _{x \rightarrow 0^{+}} \frac{1 / x}{-\frac{1}{2} x^{-3 / 2}}=-2 \lim _{x \rightarrow 0^{+}} \sqrt{x}=0 \Rightarrow \\\lim _{x \rightarrow 0^{+}} x^{\sqrt{x}}=\lim _{x \rightarrow 0^{+}} e^{\ln y}=e^{0}=1\end{array}$$

Derivatives

Differentiation

Volume

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

Okay. So for this problem we start off with the function that were given, but we can rewrite it. We can turn this into the natural log of Y equals the limit um of the natural log. We can move the natural log on the outside. So we have the natural log of the limit of our function that will allow us to rewrite things. So once we rewrite it we can get that. The natural log of why equals the limit of -2 times the natural log I have one over the square root of X Divided by one over the square root of X. The reason why we want to do this is so we can make this A. T. And now as X approaches zero from the right, he is approaching infinity. Then we'll use local tells role in order to take the derivative of the top and bottom. This will end up giving us zero. But keep in mind that we made this, the natural log of Y equals zero. So based on that we'll see that why has to equal one for the final answer.

California Baptist University

Derivatives

Differentiation

Volume

Lectures

Join Bootcamp