Question

Prove that... (LaTeX)

Answer 1

\[\Large{\lim_{x\rightarrow0^{+}}\sqrt{x}e^{\cos{\frac{1}{x}}}=0}\]

Answer 2

so far...
\[-1\le\cos{\frac{1}{x}}\le1\]\[e^{-1}\le e^{\cos{\frac{1}{x}}}\le e^{1}\]\[\sqrt{x}e^{-1}\le\sqrt{x}e^{\cos{\frac{1}{x}}}\le\sqrt{x}e^{1}\]

Answer 3

I wouldn't know the limits of these.. I think right side one with \(\Large{\sqrt{x}e^{1}}\) is this?\[\Large{\lim_{x\rightarrow0^{+}}=0}\]

Answer 4

hint:
calculate the limits of the bounds, and use squeeze theorem to find limit (if both bounds have the same value.)

Answer 5

Yeah, I don't know how to find the bound limits... @mathmate

Answer 6

Both bounds have the same limit, and they are \(0\) like you think.
\[\lim_{x\to0^+}\frac{\sqrt x}{e}=\frac{1}{e}\lim_{x\to0^+}\sqrt x=0\]\[\lim_{x\to0^+}e\sqrt x=e\lim_{x\to0^+}\sqrt x=0\]which means the limit you want to find is ...

Answer 7

Okay, I get that, but why are they 0? How did the answer come about?

Answer 8

a constant multiplied by 0 = ?

Answer 9

Do you feel the need to prove that \(\lim\limits_{x\to0^+}\sqrt x=0\)? See example 4 here: http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx

I assumed that was a fact you could use, but there's no harm in proving this limit while you're at it.

Answer 10

\(\color{blue}{\text{Originally Posted by}}\) @HolsterEmission
Do you feel the need to prove that \(\lim\limits_{x\to0^+}\sqrt x=0\)? See example 4 here:
http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx


I assumed that was a fact you could use, but there's no harm in proving this limit while you're at it.
\(\color{blue}{\text{End of Quote}}\)

I did need to see the proof, so thank you very much for that ☺