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I'm having a hell of a time understanding how to apply the Squeeze Theorem and the corresponding theorems to solving problems like the following.

$\lim_{x\to 0} \, \frac{x^2}{\sin ^2(x)}$

So I can see that this is essentially the inverse of what Rogawski refers to as Theorem 2 (2.6, if you're following along at home...)

$\lim_{x\to 0} \, \frac{\sin (\theta )}{\theta }=1$

And I'm fairly comfortable with the proof of that theorem. For the purposes of the assignment, however, I am still confused, in that I am supposed to apply this theorem to evaluate.

I have some hazy notion that I can just invert the function and so deduce that the limit is 1 because the squeezing inequality still expresses important relation even when I take the reciprocal of each term, but that's not exactly the same as understanding the process here...

Community
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Ben
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2 Answers2

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From applying the bounds on the sine function $x-x^3/6\le \sin x\le x$ we have

$\left|\frac{1}{(1-x^2/6)^2}\right|=\left|\frac{x^2}{(x-x^3/6)^2}\right|\ge \left|\frac{x^2}{\sin^2 (x)}\right|\ge \left|\frac{x^2}{x^2}\right|=1$

So, as $x \to 0$, we see that the ratio of interest is squeezed by $1$.

Mark Viola
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  • ok, but how would I select these bounding functions? – Ben Apr 15 '15 at 18:22
  • There are many ways to define the sine function. For example, one can define $\sin x$ in terms of a power series ( i.e., $\sin x= \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ ). Another way is to define the sine function as the inverse function of $\int_0^x \frac{dt}{\sqrt{1-t^2}}$, $0<x<1$. – Mark Viola Apr 15 '15 at 18:51
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$\lim_{x\to 0} [\frac{x^2}{\sin^2(x)}] =\frac{\lim_{x\to 0}[x^2]}{\lim_{x\to 0}[\sin^2(x)]} $=L
Then $ \frac{\lim_{x\to 0}[\sin^2(x)]}{\lim_{x\to 0}[x^2]}=\frac{1}{L}$
$\lim_{x\to 0}{(\frac{\sin(x)}{x})}^2$
$\lim_{x\to 0}(\frac{\sin(x)}{x})$ . $\lim_{x\to 0}{(\frac{\sin(x)}{x})}$
This function $\frac{\sin(x)}{x}$ is sandwichied between 1 and $\cos(x)$. The limit of these two functions is 1.
Thus, $\lim_{x\to 0}(\frac{\sin(x)}{x})$=1
And $\frac{1}{L}$=1 So $L=1$
Edit: For those, who are saying that limits cannot be taken as quotient here.
I have taken the limit to be L, which may not be defined, and those who know about the L'Hopital's Rule, if you see the proof of it, it also explains the thing above mentioned.
If we simply break our the Rule here-
$L=\lim_{x \to 0}\frac{\frac{d}{dx}x^2}{\frac{d}{dx}\sin^2x}$
Again we get L to be of the form $\frac{0}{0}$
So L would again equal the derivative of this fraction. Which is-
$L=\lim_{x \to 0}\frac{2}{2\cos(2x)}=1$

Aditya Agarwal
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  • The middle term of the first line is incorrect. One cannot take the limit of a quotient (which might exist) as the quotient of the limits when the limit of the denominator is zero (i.e., 0/0 is undefined). The second line is incorrect for the same reason. After that, your approach is sound. – Mark Viola Apr 15 '15 at 19:00
  • I don't think so. Because that is why I have taken it as L, I haven't given the variable L, a defined value.. – Aditya Agarwal Apr 19 '15 at 08:17