Should I use squeeze theorem? how?

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Should I use squeeze theorem? how?

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

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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
Yeah any time you see a sine function, you can usually squeeze it away since its range is always bounded. \[0 \le \sin^2(f(x)) \le 1\] In this case \[f(x)=\frac{2 \pi}{x}\] but it could literally be any real valued function, doesn't matter. Now we can take our inequality: \[-\sqrt{x}0 \le \sqrt{x}\sin^2(f(x)) \le \sqrt{x}\] And I think it might be easy to see how to continue with the squeeze theorem from here! The main thing to take away is that the bounded nature of sine or cosine really ends up at most contributing some constant to any limit, so you can kinda throw it away very easily in your mind once you get used to the idea that it won't contribute to your specific limit.
on the very first equation you wrote out, did you mean to place sin between -1 and 1? instead of 0 and 1.
Nah cause I squared it :P
Agh, wait a moment. Let me write out the original one because I think thomas forgot to put the 1+ in front of the sin in there o.o or maybe I forgot too
You have the right idea I just took your inequality that you're thinking of a step further cause \[a^2 \ge 0\]
\[\lim_{x \rightarrow 0^+}\sqrt{x}[1+\sin^2(2\pi/x)]=0\]
Actually the limit is: \[ \lim_{x\to 0+}\sqrt{x}\left(1+\sin\left(\frac{2\pi}{x}\right)\right)={\lim_{x\to 0+}\sqrt{x}}+\lim_{x\to 0+}\sqrt{x}\sin\left(\frac{2\pi}{x}\right) \] The results still apply.
..except sin^2
Apparently I had a brain fart thinking even though \(0\leq\sin^2(x)\leq1\) and \(\lim_{x\to0+}0\sqrt{x}=\lim_{x\to0+}\sqrt{x}=0\) still doesn't mean that \(\lim_{x\to0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)=0\).
But you are dropping the 1+ in front of sine!
@Empty does the thing you did still work even with the 1+? where does it go? I can pretty easily do the theorem, only that 1 is messing me up.
Maximum of sin(x) is 1. For some value of x, sin(x)=0 From the maximum of sin(x) we can deduce that the maximum of sin^2(x)=1. As for some value of x sin(x)=0 so sin^2(x)=0. Since the minimum value of x^2 is 0, the minimum of sin^2(x)=(sin(x))^2=0.
\[\lim_{x\to 0+}\sqrt{x}\left(1+\sin^2\left(\frac{2\pi}{x}\right)\right)={\lim_{x\to 0+}\sqrt{x}}+\lim_{x\to 0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)\] Both limit exist so you can expand the brackets
I had quite a lot of brain fart in this question.
\[\large\rm 0\lt \sin^2\left(\frac{2\pi}{x}\right)\le1\]\[\large\rm 1\lt 1+\sin^2\left(\frac{2\pi}{x}\right)\le2\]\[\large\rm \sqrt{x}\lt \sqrt{x}\left[1+\sin^2\left(\frac{2\pi}{x}\right)\right]\le2\sqrt{x}\]Why is the +1 messing you up nini? :o You still get zeroes on both ends of this inequality, ya?
*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
Maybe it'll make sense to look at the range of \(\sin x\): \[-1 \le \sin x \le 1\] The range is just \([-1,1]\) But when we square it, the negative part of the range goes away: \[0 \le \sin^2 x \le 1\] since \([0,1]\) is the new range.
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!
I think my inequality was wrong. For some reason I was thinking the sine function should be restricted to (0,1] because of the x in the denominator. But clearly you can put in x=2 to get get zero :P Shouldn't be strict inequality.
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!
hm, @zepdrix ?
ya :)
yaay thank you guys!

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