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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2Why? What does sin mean without Taylor series? How is it defined? – jimjim May 23 '13 at 19:33
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18By divination, the answer is $-1/3$. – Julien May 23 '13 at 19:35
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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 – mnsh May 23 '13 at 19:35
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2See this: http://math.stackexchange.com/questions/285942/find-limit-without-using-lhopital-or-taylors-series?rq=1 – Sungjin Kim May 23 '13 at 19:36
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1And what does the teacher mean by sin x? Without using Taylor series. – jimjim May 23 '13 at 19:37
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You will need an inequality as in the first answer of the link, but the inequality comes from the terms of Taylor series of $\sin x$ – Sungjin Kim May 23 '13 at 19:37
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also my teacher said there is a simple way to solve the problem . the school open after 2 day and if i have no answer he give me the answer after 2 day . notice that in secondary school in my country there is no L'Hopital or Taylor series just integral and derivative – mnsh May 23 '13 at 19:40
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That's a tough one and I have a certain inclination in what part of the world your school might be. But you say you can use derivatives except if the derivatives are in a numerator and denominator? That's odd. Now I would like to see a solution without any (anti) derivatives because I am clueless. – imranfat May 23 '13 at 19:44
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@imranfat and I will try . but if I try and know the solution .can I but the answer in my question or that is wrong ? – mnsh May 23 '13 at 19:47
4 Answers
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Let $I = \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)$, we have:
$$\begin{align} I = & \lim_{x\to 0}\left(\frac{1}{(2x)^2} - \frac{1}{\sin^2(2x)}\right)\\ = & \lim_{x\to 0} \left(\frac{1}{4 x^2} - \frac{1}{4 \sin^2 x\cos^2 x}\right)\\ \implies 4I = & \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)\\ \implies 3I = & \lim_{x\to 0}\left\{\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right) -\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)\right\}\\ =&-\lim_{x\to 0} \frac{1-\cos^2 x}{\sin^2 x\cos^2 x} = -\lim_{x\to 0}\frac{1}{\cos^2 x} = -1\\ \implies I = & -\frac{1}{3} \end{align}$$
achille hui
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7I think this proof is flawed (excepting the case we know the limit exists). – user 1591719 Nov 15 '13 at 16:30
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@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
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@Chris'ssis: However, the question asks to find the limit, which presupposes that the limit exists. – robjohn Nov 24 '13 at 18:17
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1@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 1591719 Nov 24 '13 at 18:39
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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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$x$ 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
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@hmedan.mnsh $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} f(x/a)$ for any $a \neq 0$, if the limit exists. – Zarrax May 23 '13 at 19:59
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so now I have an idea to the solution of last limit you wrote it . can I put it or that is wrong in the term of this site ? – mnsh May 23 '13 at 19:59
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@hmedan.mnsh I don't know if there is a rule against that, but it's ok with me personally – Zarrax May 23 '13 at 20:02
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1@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. – jimjim May 23 '13 at 20:07
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sorry about to answer my Question but Zarrax's way and her comment lead me to the answer $$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}$$
$$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}$$
$$L=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}$$
$$L=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}=\frac{3L}{27}-\frac{4}{27}$$
$$L=\frac{3L}{27}-\frac{4}{27}$$
$$L=-\frac{1}{6}$$
$$2L=-\frac{1}{3}=\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
thanks alot Zarrax thanks for every one
mnsh
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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 06:16
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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 $$ \begin{align} \lim_{x\to0}\left(\frac1{\vphantom{()^2}x^2}-\frac1{\sin^2(x)}\right) &=-\frac16\cdot2\cdot1^2\\ &=-\frac13\tag{4} \end{align} $$
