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Babynini
 one year ago
Should I use squeeze theorem? how?
Babynini
 one year ago
Should I use squeeze theorem? how?

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Prove that lim (as x approaches 0 from the positive side) sq{x}[1+sin^2(2pi/)]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3Is this the limit you want? I type the previous limit wrong. \[ \lim_{x\to0+}\sqrt{x}\sin^2(2\pi) \]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Correct, except sin^2(2pi/x) at the end there

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3This one? \[ \lim_{x\to0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right) \]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0ooh after the end there is an = 0

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Sorry sorry. So all of that that you just wrote out = 0

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0on the very first equation you wrote out, did you mean to place sin between 1 and 1? instead of 0 and 1.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Nah cause I squared it :P

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Agh, 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

Empty
 one year ago
Best ResponseYou've already chosen the best response.1You have the right idea I just took your inequality that you're thinking of a step further cause \[a^2 \ge 0\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0^+}\sqrt{x}[1+\sin^2(2\pi/x)]=0\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3Actually 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.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3Apparently 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 \(\lim_{x\to0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)=0\).

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0But you are dropping the 1+ in front of sine!

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0@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.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3Maximum 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.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3\[\lim_{x\to 0+}\sqrt{x}\left(1+\sin^2\left(\frac{2\pi}{x}\right)\right)={\lim_{x\to 0+}\sqrt{x}}+\lim_{x\to 0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)\] Both limit exist so you can expand the brackets

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3I had quite a lot of brain fart in this question.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm 0\lt \sin^2\left(\frac{2\pi}{x}\right)\le1\]\[\large\rm 1\lt 1+\sin^2\left(\frac{2\pi}{x}\right)\le2\]\[\large\rm \sqrt{x}\lt \sqrt{x}\left[1+\sin^2\left(\frac{2\pi}{x}\right)\right]\le2\sqrt{x}\]Why is the +1 messing you up nini? :o You still get zeroes on both ends of this inequality, ya?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3*Both limits Plural singular brain fart.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0ooh it all just moves up by 1?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0hahah @thomas5267 you and I both! I'm not sure why this problem is being so troublesome xD

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Maybe 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.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0And then that range [0,1] becomes [1,2] because of the 1+ in the equation in front of sine?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0(Yes, thank you, that made it make so much more sense xD)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah looks perfect to me!

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Actually one tiny thing, \[0 \le \sin^2x\] since they can be equal!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0or to get* zero, as opposed to get getting zero _ dat typo

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Ah okay, so i just need to add the "=" to each of the "<" on the left side!
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