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How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$

without using L'Hospital or Taylor series?

thanks :)

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1  
Use Taylor expansion. –  Your Ad Here Oct 19 '12 at 20:20
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... without the Axiom of Choice, without using the letter 'a', without a net... –  rschwieb Oct 19 '12 at 20:41
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GH Hardy "Pure Mathematics" Examples XLVI (page 237 in Tenth Edition) has some nice exercises starting with $x-\sin x$ and $\tan x-x$ are increasing for suitable small positive $x$ –  Mark Bennet Oct 19 '12 at 20:42
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@Iuli Perhaps because the question was changed to exclude a possibility after an answer was given - which was, I think, intended to invite alternative answers, but misguided (rude) as a way of doing so. The person who put the answer has had to amend it to explain why it was given. Instead you might have put - "there is a nice Taylor Series answer already, but I'm also looking for other ideas and ways of attacking this problem". I was not the down voter. –  Mark Bennet Oct 19 '12 at 21:21
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See in particular this answer, which also covers the case with $\sin$ instead of $\tan$: math.stackexchange.com/a/158134/1242 –  Hans Lundmark Oct 20 '12 at 13:54

4 Answers 4

up vote 25 down vote accepted

Let $L = \lim_{x \to 0} \dfrac{x - \sin(x)}{x^3}$. We then have \begin{align} L & = \underbrace{\lim_{y \to 0} \dfrac{3y - \sin(3y)}{27y^3} = \lim_{y \to 0} \dfrac{3y - 3\sin(y) + 4 \sin^3(y)}{27y^3}}_{\sin(3y) = 3 \sin(y) - 4 \sin^3(y)}\\ & = \lim_{y \to 0} \dfrac{3y - 3\sin(y)}{27 y^3} + \dfrac4{27} \lim_{y \to 0} \dfrac{\sin^3(y)}{y^3} = \dfrac{3}{27} L + \dfrac4{27} \end{align} This gives us $24L = 4 \implies L = \dfrac16$

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2  
Thanks for ending this madness :D. –  Your Ad Here Oct 19 '12 at 20:49
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Dont you need first to prove that the limit exists? For example $\lim_ {x \rightarrow \infty} \cos (x)=\lim_ {x \rightarrow \infty} \cos (2x)=\lim_ {x \rightarrow \infty} \cos (x) ^2-1\Rightarrow a=2a^2-1$ and solve –  clark Oct 19 '12 at 20:53
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How do you show that the $\lim$ exists in the firs place? –  Hagen von Eitzen Oct 19 '12 at 20:53
    
Yes, that is a good point. Let me think of a method to prove that limit exists and then add it to the post. –  user17762 Oct 19 '12 at 20:55
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It exists by l'Hopital's rule, or the Taylor expansion. :) Seriously, though, assuming you can prove the existing somewhere, that is a really great idea I would never have thought of. –  Graphth Oct 19 '12 at 21:22

Short answer: function discussion. Long answer (details) is here.

First step. Your fraction is an even function so it is enough to consider the right limit.

Second step. Define the functions $f(x):=\sin(x)-x+\frac{1}{6}x^3$ and $g(x):=\sin(x)-x+\frac{1}{6}x^3-\frac{1}{120}x^5$ on the interval $[0, 0.1]$.

Third step. Prove that $f$ is strictly increasing and $g$ is strictly decreasing. ($f(0)=0$, it is enough $f'>0$. $f'(0)=0$, so it is enough $f''>0$ and so on. Similarly for $g$.

Forth step. From the third step $$ x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5. $$ The limit follows.

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Note: The original question didn't say anything about not using Taylor series. Then, after I answered this, the OP changed the question and said didn't want Taylor series either.

Another option would be using power series (aka Maclaurin series aka Taylor series centered at 0)

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

At this point, it's pretty easy.

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yes, yes . another method ? –  Iuli Oct 19 '12 at 20:21
    
This is the definition of $\sin$, so I any other method would necessarily be derived from this. –  Your Ad Here Oct 19 '12 at 20:24
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@IHaveAStupidQuestion That is one definition of $\sin$. There are others. –  Pedro Tamaroff Oct 19 '12 at 20:26
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@Argon The first definition is the one used in high school. The second definition is literally the same as the Taylor expansion by using the definition of exp. –  Your Ad Here Oct 19 '12 at 20:33
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You could make up arbitrarily many nice formulas that are equal to $sin(x)$. I think this discussion is pointless. –  Your Ad Here Oct 19 '12 at 20:35

Assuming $f$ is sufficiently smooth, repeated application of the fundamental theorem of the calculus gives (finite Taylor expansion) $$f(x) = f(0)+f'(0)x+\frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \int_0^x (x-t)^2 \frac{(f'''(t)-f'''(0))}{2!}\, dt$$

Using the fact that $\sin' = \cos, \cos' = -\sin$, we can then expand $f(x) = \sin x$ as $$ \sin x = x -\frac{x^3}{6} + \int_0^x (x-t)^2 \frac{(-\cos t +1)}{2!}\, dt$$ Let $\epsilon>0$, and choose $\delta>0$ such that if $|t|< \delta$, then $|1-\cos t| < \epsilon$. Then, replacing $1-\cos t$ by $\epsilon$ and integrating, we have the estimate $$ |\sin x - x + \frac{x^3}{6} | \leq \epsilon \frac{|x|^3}{6}$$ If $0 < |x| < \delta$, then dividing through by $|x^3|$ gives: $$ | \frac{\sin x - x}{x^3} +\frac{1}{6} | \leq \frac{\epsilon}{6} < \epsilon$$ The desired limit follows.

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