Question

Should I use squeeze theorem? how?

Answer 1

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

Answer 2

@Empty :)

Answer 3

@zepdrix

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Yep!

Answer 8

ooh after the end there is an = 0

Answer 9

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

Answer 10

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.

Answer 11

on the very first equation you wrote out, did you mean to place sin between -1 and 1? instead of 0 and 1.

Answer 12

Nah cause I squared it :P

Answer 13

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

Answer 14

You have the right idea I just took your inequality that you're thinking of a step further cause \[a^2 \ge 0\]

Answer 15

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

Answer 16

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.

Answer 17

..except sin^2

Answer 18

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 \(\lim_{x\to0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)=0\).

Answer 19

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

Answer 20

@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.

Answer 21

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.

Answer 22

\[\lim_{x\to 0+}\sqrt{x}\left(1+\sin^2\left(\frac{2\pi}{x}\right)\right)={\lim_{x\to 0+}\sqrt{x}}+\lim_{x\to 0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)\]
Both limit exist so you can expand the brackets

Answer 23

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

Answer 24

\[\large\rm 0\lt \sin^2\left(\frac{2\pi}{x}\right)\le1\]\[\large\rm 1\lt 1+\sin^2\left(\frac{2\pi}{x}\right)\le2\]\[\large\rm \sqrt{x}\lt \sqrt{x}\left[1+\sin^2\left(\frac{2\pi}{x}\right)\right]\le2\sqrt{x}\]Why is the +1 messing you up nini? :o
You still get zeroes on both ends of this inequality, ya?

Answer 25

*Both limits

Plural singular brain fart.

Answer 26

ooh it all just moves up by 1?

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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

Answer 31

is that..correct?

Answer 32

Yeah looks perfect to me!

Answer 33

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

Answer 34

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.

Answer 35

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

Answer 36

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

Answer 37

hm, @zepdrix ?

Answer 38

ya :)

Answer 39

yaay thank you guys!