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One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?

$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$

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5 Answers 5

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Here's a way which avoids derivatives and integrals.

Assume that we know that $\frac{\sin x}{x} \to 1$ as $x \to 0$. Then we also know that $\frac{1-\cos x}{x^2} = \frac12 \left( \frac{\sin(x/2)}{x/2} \right)^2 \to \frac12$.

Now, $$ \frac{\tan x - x}{x^3} = \frac{1}{\cos x} \left( \frac{\sin x - x}{x^3} + \frac{1-\cos x}{x^2} \right), $$ so we are done if we can compute $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac16$. The reason that I rewrote it like this is that I was asked by a colleague about ten years ago whether that limit could be done in an elementary way. :-) I came up with the following:

Let $$ f(x) = \frac{x - \sin x}{x^3} = \frac{1 - \frac{\sin x}{x}}{x^2}. $$ (Here I've changed the sign so that the limit will be positive.) Since $f$ is an even function, it's enough to consider $x>0$.

Fix a positive integer $n$. To begin with, we have $$ x = 2^n \frac{x}{2^n} > 2^n \sin \frac{x}{2^n}. $$ (I'm assuming that we also know that $0 < \sin x < x < \tan x$ for $0 < x < \pi/2$.) Multiply this inequality by $\prod_{k=1}^n \cos\frac{x}{2^k}$ and use the double angle formula repeatedly, as follows (illustrated for the case $n=3$): $$ \begin{split} x \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2} & > 2^3 \sin\frac{x}{8} \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2} \\ & = 2^2 \sin\frac{x}{4} \cos\frac{x}{4} \cos\frac{x}{2} \\ & = 2^1 \sin\frac{x}{2} \cos\frac{x}{2} \\ & = \sin x. \end{split} $$ This implies (again for $n=3$, but the general pattern is hopefully clear) $$ \begin{split} 1 - \frac{\sin x}{x} & > 1 - \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2} \\ & = \left( 1 - \cos\frac{x}{8} \right) + \cos\frac{x}{8} \left( 1 - \cos\frac{x}{4} \right) + \cos\frac{x}{8} \cos\frac{x}{4} \left( 1 - \cos\frac{x}{2} \right). \end{split} $$ We know that $\frac{1 - \cos(x/2^k)}{x^2} = \frac{1 - \cos(x/2^k)}{2^{2k} (x/2^k)^2} \to \frac{1}{2^{2k+1}}$, so after dividing this inequality by $x^2$ we find in the limit (for general $n$) that $$ \liminf_{x \to 0^+} f(x) \ge \sum_{k=1}^n \frac{1}{2^{2k+1}} = \frac16 \left( 1 - \frac{1}{4^n} \right). $$ This holds for every $n$, hence $$ \liminf_{x \to 0^+} f(x) \ge \frac16. $$

The other direction is similar. Start with $$ x = 2^n \frac{x}{2^n} < 2^n \tan\frac{x}{2^n} = 2^n \frac{\sin(x/2^n)}{\cos(x/2^n)}. $$ This leads to $$ \begin{split} 1 - \frac{\sin x}{x} & < 1 - \cos\frac{x}{2^n} \cdot (\text{same product of cosines as above}) \\ & = 1 - \cos\frac{x}{2^n} + \cos\frac{x}{2^n} \cdot (1 - (\text{that product})) \\ & = 1 - \cos\frac{x}{2^n} + \cos\frac{x}{2^n} \cdot (\text{same expression as above}). \end{split} $$ Divide by $x^2$ and let $x \to 0^+$: $$ \limsup_{x \to 0^+} f(x) \le \frac{1}{2^{2n+1}} + \frac16 \left( 1 - \frac{1}{4^n} \right). $$ Let $n \to \infty$: $$ \limsup_{x \to 0^+} f(x) \le \frac16. $$ It follows that $\lim_{x \to 0^+} f(x) = \frac16$, and therefore by symmetry $\lim_{x \to 0} f(x) = \frac16$,which is what we wanted to show.

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  • $\begingroup$ Another nice way of doing this is in this answer by @robjohn (building on this answer where the heavy work is done). $\endgroup$ Jul 11, 2017 at 9:37
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There's the Cauchy or extended mean value theorem which says that ${\displaystyle {f(x) - f(y) \over g(x) - g(y)} = {f'(c) \over g'(c)}}$ for some $c$ between $x$ and $y$. You can apply it here with $f(x) = \tan(x) - x$ and $g(x) = x^3$, and you get that for some $0 < c < x$ you have $$\frac{\tan(x) - x}{x^3} = {\sec^2(c) - 1 \over 3c^2}$$ $$= {\tan^2(c) \over 3c^2}$$ $$ = {1 \over 3}{1 \over \cos^2(c)}{\sin^2(c) \over c^2}$$ Now take limits as $x$ goes to zero; $c$ goes to zero and the limit is $1/3$. This is all somewhat tongue in cheek of course, since you can get L'Hopital pretty quickly from the extended mean value theorem, but it does satisfy your request to not use it or Taylor polynomials :)

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  • $\begingroup$ that's right. Thanks. I was thinking of some geometrical approach. $\endgroup$ Jun 13, 2012 at 19:44
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Encouraged by Hans Lundmark's answer, I'm posting my own solution without derivatives and integrals.

The triple-angle formula for $\tan$ is $$\tan 3\theta = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.$$ Suppose $\lim_{x\to0}(\tan x-x)/x^3 = c$. Letting $x = 3\theta$, we then have $$\begin{align} c &= \lim_{x\to0} \frac{\tan x-x}{x^3} \\ &= \lim_{\theta\to0} \frac{\tan 3\theta-3\theta}{27\theta^3} \\ &= \lim_{\theta\to0} \frac{3\tan\theta - \tan^3\theta-3\theta+9\theta\tan^2\theta}{27\theta^3(1-3\tan^2\theta)} \end{align}$$ We can get rid of $1/(1-3\tan^2\theta)$ because its limit is $1$. Next we start pulling out terms and find $$\begin{align} c &= \lim_{\theta\to0}\frac{3\tan\theta-3\theta}{27\theta^3} - \lim_{\theta\to0}\frac{\tan^3\theta}{27\theta^3} + \lim_{\theta\to0}\frac{9\theta\tan^2\theta}{27\theta^3} \\ &= \frac19c - \frac1{27} + \frac{1}{3}, \end{align}$$ because $\lim_{\theta\to0}(\tan\theta)/\theta = 1$. So $8c/9 = 8/27$, or $c = 1/3$.

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  • $\begingroup$ Nice. Similar method to mine! $\endgroup$
    – user17762
    Jun 14, 2012 at 7:07
  • $\begingroup$ Ah, much simpler than my answer! Except that you use heavy machinery like the triple angle formula and I only used the double angle formula. ;-) $\endgroup$ Jun 14, 2012 at 7:59
  • $\begingroup$ @Rahul Narain: nice answer! Thanks. $\endgroup$ Jun 14, 2012 at 8:08
  • $\begingroup$ @Hans: Yes, I pulled out the big guns for this one! :) Actually, I just didn't realize that the double angle formula works fine with this approach -- it becomes equivalent to Marvis' answer. $\endgroup$
    – user856
    Jun 14, 2012 at 9:21
  • $\begingroup$ There's a problem with the existence of the limit, which has to be shown before this argument is complete. See my comment to Marvis's answer. $\endgroup$ Jun 14, 2012 at 14:10
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Here is a different approach. Let $$L = \lim_{x \to 0} \dfrac{\tan(x) - x}{x^3}$$ Replacing $x$ by $2y$, we get that \begin{align} L & = \lim_{y \to 0} \dfrac{\tan(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2 \tan(y) + 2 \tan(y) - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \dfrac{\dfrac{2 \tan^3(y)}{1 - \tan^2(y)} + 2 \tan(y) - 2y}{(2y)^3}\\ & = \lim_{y \to 0} \left(\dfrac{2 \tan^3(y)}{8y^3(1 - \tan^2(y))} + \dfrac{2 \tan(y) - 2y}{8y^3} \right)\\ & = \lim_{y \to 0} \left(\dfrac{2 \tan^3(y)}{8y^3(1 - \tan^2(y))} \right) + \lim_{y \to 0} \left(\dfrac{2 \tan(y) - 2y}{8y^3} \right)\\ & = \dfrac14 \lim_{y \to 0} \left(\dfrac{\tan^3(y)}{y^3} \dfrac1{1 - \tan^2(y)} \right) + \dfrac14 \lim_{y \to 0} \left(\dfrac{\tan(y) - y}{y^3} \right)\\ & = \dfrac14 + \dfrac{L}4 \end{align} Hence, $$\dfrac{3L}{4} = \dfrac14 \implies L = \dfrac13$$

EDIT

In Hans Lundmark answer, evaluating the desired limit boils down to evaluating $$S=\lim_{x \to 0} \dfrac{\sin(x)-x}{x^3}$$ The same idea as above can be used to evaluate $S$ as well. Replacing $x$ by $2y$, we get that \begin{align} S & = \lim_{y \to 0} \dfrac{\sin(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y) - 2y}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y) - 2 \sin(y) + 2 \sin(y) - 2y}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2 \sin(y) - 2y}{8y^3} + \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y)-2 \sin(y)}{8y^3}\\ & = \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) - y}{y^3} - \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) (1 - \cos(y))}{y^3}\\ & = \dfrac{S}4 - \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) 2 \sin^2(y/2)}{y^3}\\ & = \dfrac{S}4 - \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 - \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \lim_{y \to 0} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 - \dfrac18\\ \dfrac{3S}4 & = - \dfrac18\\ S & = - \dfrac16 \end{align}

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    $\begingroup$ Huh, I had assumed that to get $\tan$ to third order one would have to use the triple angle formula. I guess I was mistaken. $\endgroup$
    – user856
    Jun 14, 2012 at 7:11
  • $\begingroup$ @Marvis: very elegant! Thanks! $\endgroup$ Jun 14, 2012 at 8:05
  • $\begingroup$ @Marvis: this proof also works for showing that $\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^2} = 0$. $\endgroup$ Jun 14, 2012 at 8:16
  • $\begingroup$ @Chris Yes. This is powerful neat little trick. I learnt this trick in my last year at high school and yes you can show that $\lim_{x \to 0} \dfrac{\tan(x)-x}{x^2} =0$ using this trick as well. $\endgroup$
    – user17762
    Jun 14, 2012 at 8:23
  • $\begingroup$ @Marvis: i've just noticed now that it works great and very simple by only using: $\sin(x)<x<\tan(x),\space 0< x <\frac{\pi}{2}.$. It's nice to have a lot of tackling ways for each problem. :-) $\endgroup$ Jun 14, 2012 at 10:16
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The statement $\dfrac{\tan(x)-x}{x^3} \to c$ as $x \to 0$ is equivalent to $\tan(x) = x + c x^3 + o(x^3)$ as $x \to 0$, so this is a statement about a Taylor polynomial of $\tan(x)$, and I'm not sure what would count as doing that "without Taylor". However, one thing you could do is start from $$\sin(x) = x + o(x)$$ integrate to get $$\cos(x) = 1 - x^2/2 + o(x^2)$$ then $$\sec(x) = \frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$$ $$\sec^2(x) = \left(1 + x^2/2 + o(x^2)\right)^2 = 1 + x^2 + o(x^2)$$ and integrate again to get $$\tan(x) = x + x^3/3 + o(x^3)$$

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  • $\begingroup$ thanks. Your approach is nice and interesting. Initially my first thought was to find some geometrical approch, but it doesn't seem that easy.I'm still trying to find a link between this limit and the intersection point of medians in a triangle. $\endgroup$ Jun 13, 2012 at 19:48

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