## anonymous one year ago Review: Limits Evaluate. Review: Limits Evaluate. $$\sf \Large lim_{x \rightarrow 0^+} \sqrt{x}e^{sin(\frac{\pi}{x})}$$

1. anonymous

I assume that since $$\sf \large lim_{x \rightarrow 0^+} \sqrt{x}$$ is zero, this limit will be equal to zero too. Because anything that we multiply with zero is zero and the limit law says that $$\sf lim_{x \rightarrow a} [f(x)g(x)]=lim_{x \rightarrow a} f(x) •lim_{x \rightarrow a} g(x)$$ so for my question: $$\sf = lim_{x \rightarrow 0^+} \sqrt{x} • lim_{x \rightarrow 0^+} e^{sin(\frac{\pi}{x})}\\= 0 \ •\ lim_{x \rightarrow 0^+} e^{sin(\frac{\pi}{x})}$$ i know that pi/x will be undefined and my reasoning is not that good, so yeah I need help with explaining it.

2. ganeshie8

are you allowed to use L'hopital ?

3. Astrophysics

I think squeeze theorem

4. zepdrix

ya squeeze seems nice for this one :)

5. anonymous

l'hopital's rule is not yet allowed to use in this problem

6. Astrophysics

Since $- 1 \le \sin \left( \frac{ \pi }{ x } \right) \le 1$

7. zepdrix

$\large\rm -1\le \sin\left(\frac{\pi}{x}\right)\le1$Exponentiate everything,$\large\rm e^{-1}\le e^{\sin\left(\frac{\pi}{x}\right)}\le e^{1}$Throw some of that square root magic into there,$\large\rm \sqrt x e^{-1}\le \sqrt x e^{\sin\left(\frac{\pi}{x}\right)}\le \sqrt x e^{1}$

8. zepdrix

And then let x approach 0, for the left and right most sides, ya? :o

9. Astrophysics

Looks good!

10. anonymous

oh okay,got it! :D Thank you so much!

11. zepdrix

yay team \c:/