Question

Review: Limits

Evaluate.
Review: Limits
Evaluate.
\(\sf \Large lim_{x \rightarrow 0^+} \sqrt{x}e^{sin(\frac{\pi}{x})}\)

Answer 1

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.

Answer 2

are you allowed to use L'hopital ?

Answer 3

I think squeeze theorem

Answer 4

ya squeeze seems nice for this one :)

Answer 5

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

Answer 6

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

Answer 7

\[\large\rm -1\le \sin\left(\frac{\pi}{x}\right)\le1\]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,\[\large\rm \sqrt x e^{-1}\le \sqrt x e^{\sin\left(\frac{\pi}{x}\right)}\le \sqrt x e^{1}\]

Answer 8

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

Answer 9

Looks good!

Answer 10

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

Answer 11

yay team \c:/