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Prove that lim (as x approaches 0 from the positive side)
sq{x}[1+sin^2(2pi/)]

Is this the limit you want? I type the previous limit wrong.
\[
\lim_{x\to0+}\sqrt{x}\sin^2(2\pi)
\]

Correct, except sin^2(2pi/x) at the end there

This one?
\[
\lim_{x\to0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)
\]

Yep!

ooh after the end there is an = 0

Sorry sorry.
So all of that that you just wrote out = 0

Nah cause I squared it :P

\[\lim_{x \rightarrow 0^+}\sqrt{x}[1+\sin^2(2\pi/x)]=0\]

..except sin^2

But you are dropping the 1+ in front of sine!

I had quite a lot of brain fart in this question.

*Both limits
Plural singular brain fart.

ooh it all just moves up by 1?

hahah @thomas5267 you and I both! I'm not sure why this problem is being so troublesome xD

And then that range [0,1] becomes [1,2] because of the 1+ in the equation in front of sine?

(Yes, thank you, that made it make so much more sense xD)

is that..correct?

Yeah looks perfect to me!

Actually one tiny thing, \[0 \le \sin^2x\] since they can be equal!

or to get* zero,
as opposed to get getting zero -_- dat typo

Ah okay, so i just need to add the "=" to each of the "<" on the left side!

ya :)

yaay thank you guys!