💬 👋 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.$\displaystyle \lim_{x\to \infty} x\sin(\pi /x)$

## This limit has the form $\infty \cdot 0 .$ We'll change it to the form $\frac{0}{0}$$\lim _{x \rightarrow \infty} x \sin (\pi / x)=\lim _{x \rightarrow \infty} \frac{\sin (\pi / x)}{1 / x} \stackrel{\Perp}{=} \lim _{x \rightarrow \infty} \frac{\cos (\pi / x)\left(-\pi / x^{2}\right)}{-1 / x^{2}}=\pi \lim _{x \rightarrow \infty} \cos (\pi / x)=\pi(1)=\pi$

Derivatives

Differentiation

Volume

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

Mhm. The first problem we're gonna be taking the limit as X goes to infinity of the following function. That's gonna be X. Sine of pi Uber X. So X sign of high over X is going to be our function. Okay? So when we substitute one over X for thi this is just gonna be pi T. And we know that as X goes to infinity to equal zero. So we can just change how we view the limit. So in this case now um this is gonna be a sign of pirate T over T. When T goes to zero now we have 0/0. That's an indeterminate form. So we'll take the uh derivative of the numerator. That'll be high co sign right? And this will just be one on the bottom. Then what will end up getting as a result is high time to co sign of zero which is one Over one. So our final answer will be high.

California Baptist University

Derivatives

Differentiation

Volume

Lectures

Join Bootcamp