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Find the limit without the use of L'Hôpital's rule or Taylor series

$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$

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Why? What does sin mean without Taylor series? How is it defined? – Arjang May 23 '13 at 19:33
By divination, the answer is $-1/3$. – 1015 May 23 '13 at 19:35
my teacher give that limit for me and tell me there is a methode without use L'Hopital's rule or Taylor series . because of that i try to find any way without L'Hopital's rule or Taylor series – mhd.math May 23 '13 at 19:35
See this:… – i707107 May 23 '13 at 19:36
And what does the teacher mean by sin x? Without using Taylor series. – Arjang May 23 '13 at 19:37
up vote 19 down vote accepted

Let $I = \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)$, we have:

$$\begin{align} I = & \lim_{x\to 0}\left(\frac{1}{(2x)^2} - \frac{1}{\sin^2(2x)}\right)\\ = & \lim_{x\to 0} \left(\frac{1}{4 x^2} - \frac{1}{4 \sin^2 x\cos^2 x}\right)\\ \implies 4I = & \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)\\ \implies 3I = & \lim_{x\to 0}\left\{\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right) -\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)\right\}\\ =&-\lim_{x\to 0} \frac{1-\cos^2 x}{\sin^2 x\cos^2 x} = -\lim_{x\to 0}\frac{1}{\cos^2 x} = -1\\ \implies I = & -\frac{1}{3} \end{align}$$

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I think this proof is flawed (excepting the case we know the limit exists). – user 1618033 Nov 15 '13 at 16:30
@Chris'ssis: I agree. Proofs of this nature can be used to show that $\lim_{x\to0}\sin(2\pi\log_2(x))=0$ since $\sin(2\pi\log_2(2x))=\sin(2\pi\log_2(x))$. – robjohn Nov 24 '13 at 17:48
@Chris'ssis: However, the question asks to find the limit, which presupposes that the limit exists. – robjohn Nov 24 '13 at 18:17
@robjohn Well, in my opinion, "Find the limit" may also lead you to the fact the limit doesn't exist. In my math books, there are situations where the limits don't exist, although the text is written like that "Find the limits". – user 1618033 Nov 24 '13 at 18:39

Well, you have $$\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right) = \left(\frac{1}{x} - \frac{1}{\sin x}\right)\left(\frac{1}{x} + \frac{1}{\sin x}\right)$$ $$=\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)\left(1+ \frac{x}{\sin x}\right)$$ Since the limit of $\left(1+ \frac{x}{\sin x}\right)$ is $2$, your answer will be $$2\lim_{x \rightarrow 0}\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)$$ $$=2\lim_{x \rightarrow 0}\left(\frac{x}{\sin x}\right)\left(\frac{\sin x - x}{x^3}\right)$$ $$=2\lim_{x \rightarrow 0}\frac{\sin x - x}{x^3}$$ An elementary calculus way of showing this limit is $-{1 \over 6}$ can be found here.

So the overall answer is $-\frac{1}{3}$.

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What is sin x without using Taylor? – Arjang May 23 '13 at 19:50
Defined geometrically through circles – Zarrax May 23 '13 at 19:51
And the limit? How is that defined geometrically? – Arjang May 23 '13 at 19:52
$x$ is the angle, $\sin x$ is the sine of that angle. So you have to show the limit exists as the angle tends to zero, and find out what the limit equals. – Zarrax May 23 '13 at 19:53
@hmedan.mnsh : without Taylor sin x is meaningless in the context of taking limits without doing some pretty looking geometrical figures, the last limit is just as hard without using Taylor. – Arjang May 23 '13 at 20:07

sorry about to answer my Question but Zarrax's way and her comment lead me to the answer $$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}$$

$$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}$$

$$L=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}$$

$$L=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}=\frac{3L}{27}-\frac{4}{27}$$



$$2L=-\frac{1}{3}=\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$

thanks alot Zarrax thanks for every one

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Again, this requires a proof that the limit exists. The line $L = {3 \over 27}L - {4 \over 27}$ is also satisfied by $L = \infty$. – user164587 Sep 21 '15 at 6:16

We will use that $\lim\limits_{x\to0}\frac{\sin(x)}{x}=1$, which can be shown geometrically.

First note that $$ \begin{align} \frac1{\vphantom{()^2}x^2}-\frac1{\sin^2(x)} &=\frac{\sin^2(x)-x^2}{x^2\sin^2(x)}\\ &=\frac{\sin^2(x)-x^2}{x^4}\left(\frac{\sin(x)}{x}\right)^{-2}\\ &=\frac{\sin(x)-x}{x^3}\left(\frac{\sin(x)}{x}+1\right)\left(\frac{\sin(x)}{x}\right)^{-2}\tag{1} \end{align} $$ Next, we have $$ \begin{align} \frac{\sin(x/2^{k-1})-2\sin(x/2^k)}{x^3} &=\frac{2\sin(x/2^k)(\cos(x/2^k)-1)}{x^3}\\ &=\frac{\sin(x/2^k)}{x^3}\frac{2(\cos^2(x/2^k)-1)}{\cos(x/2^k)+1}\\ &=-2^{-3k}\left(\frac{\sin(x/2^k)}{x/2^k}\right)^3\frac2{\cos(x/2^k)+1}\tag{2} \end{align} $$ Therefore, $$ \begin{align} \lim_{x\to0}\frac{\sin(x)-x}{x^3} &=\lim_{x\to0}\sum_{k=1}^\infty2^{k-1}\frac{\sin(x/2^{k-1})-2\sin(x/2^k)}{x^3}\\ &=\lim_{x\to0}\sum_{k=1}^\infty-2^{-2k-1}\left(\frac{\sin(x/2^k)}{x/2^k}\right)^3\frac2{\cos(x/2^k)+1}\\ &=\sum_{k=1}^\infty-2^{-2k-1}\\ &=-\frac16\tag{3} \end{align} $$ Plugging $(3)$ into $(1)$, we get $$ \begin{align} \lim_{x\to0}\left(\frac1{\vphantom{()^2}x^2}-\frac1{\sin^2(x)}\right) &=-\frac16\cdot2\cdot1^2\\ &=-\frac13\tag{4} \end{align} $$

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