I have a problem with solution of this limit: $$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}$$ Of course, it's a very easy to solve, using (twice) L'Hôpital's rule, but I need to find out, how to do this without this rule.

I stuck in this point: $$\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$$ Everything I need to know is how to eliminate $\frac{\sin^2{x}}{x^2}$, because - as my tutor said - I can't simply substitute $1$ for this expression.

Thanks for help.

PS: It's not a homework. My tutor showed this problem as a puzzle and said, that it would be a good exercise to solve this without L'Hôpital's rule.

EDIT: Here is a way I got to the point $\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$:

\begin{align*} \lim_{x->0}{\frac{\tan{x}-x}{x^2}} &= \lim_{x->0}{\frac{(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}-x)\cdot (\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}{x^2(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}}\\ &= \lim_{x->0}{\frac{\frac{\sin^2{x}}{x^2\cos^2{x}}-1}{x(\frac{\sin{x}}{x}\frac{1}{\cos{x}}+1)}}= \lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}} \end{align*}

  • $\begingroup$ Just out of curiosity, how did you get to that limit? $\endgroup$ – Arturo Magidin Dec 16 '10 at 21:40
  • $\begingroup$ In the third term of the last line, a cosine needs a square. $\endgroup$ – Rasmus Dec 16 '10 at 22:33
  • $\begingroup$ @Rasmus: I think, it's correct, because $\cos{x}=\pm\sqrt{1-\sin^2{x}}$. But maybe I just can't see the obvious, I'm little tired of trying to solve this problem... $\endgroup$ – guram Dec 16 '10 at 22:42
  • $\begingroup$ In the numerator you have two of those. $\endgroup$ – Rasmus Dec 16 '10 at 22:45
  • $\begingroup$ @Rasmus: Of course You are right, I fix that mistake. Thank You. $\endgroup$ – guram Dec 16 '10 at 22:49

Here is an attempt at a geometric proof.

alt text

(Figure thanks to J.M)

Consider $\triangle ABC$ such that $\angle{BCA} = x$. Let $BC=1$ and so $AB = \tan x$.

Let $BE$ be the arc of radius 1 and angle $x$ drawn with $C$ as the center (note that $E$ is on AC, between $A$ and $D$ and is kind of hidden in the brown region). Note that $CE = 1$.

Now the area of the gray region is $\dfrac{\tan x}{2} - \dfrac{x}{2}$ (area of $\triangle ABC$ - area of the sector $CBE$).

Let $D$ be the perpendicular on the hypotenuse $AC$ from $B$. It can be seen that $CD = \cos x$ and thus distance from $D$ to $C$ is less than distance from $E$ to $C$ (which is $1$).

Thus the area of the gray region is less than the area of $\triangle BAD$ (gray + brown).

Now $AD = \dfrac{\sin^2 x}{\cos x}$ and thus we have that

$$ 0 < \dfrac{\tan x - x}{2} \lt \dfrac{\sin^3 x}{2\cos x}$$

And so

$$ 0 < \dfrac{\tan x - x}{2x^2} \lt \dfrac{\sin^3 x}{2x^2 \cos x}$$

Since we know that $\lim_{x \to 0+} \dfrac{\sin x}{x} = 1$, and that $\dfrac{\tan x - x}{x^2}$ is an odd function, that the limit is $0$, follows.

Previous answer, which was a feeble attempt at being pedantic:

For a way to find the limit without using more advanced concepts like McLaurin series etc...


$$\dfrac{\tan(2x) - 2x}{(2x)^2} = \dfrac{ \frac{2\tan x}{1-\tan^2 x} - 2x}{4x^2}$$

$$ = \dfrac{(2\tan x - 2x) + 2x \tan^2 x}{4x^2(1 - \tan^2 x)} = (\dfrac{\tan x - x}{2x^2} + \dfrac{x\tan^2 x}{2x^2}) \dfrac{1}{1-\tan^2 x}$$

Therefore, taking limits as $\displaystyle x \to 0$

$$ L = (\dfrac{L}{2} + 0)\dfrac{1}{1-0}$$


$$L = 0$$

There is one problem with the above, though. Can you tell what that is?

(Or rather more simply, replace $\displaystyle x$ with $\displaystyle -x$)

  • $\begingroup$ @Moron: I am not sure what problem do You mean? I think, Your solution is not only simple but also very clever, and I can't see any gap there. $\endgroup$ – guram Dec 16 '10 at 22:27
  • $\begingroup$ @guram: The above does not show that the limit is $0$. What it shows is that if the limit exists, then it is $0$. Existence of the limit needs to be shown... $\endgroup$ – Aryabhata Dec 16 '10 at 22:28
  • $\begingroup$ @Moron: But how to show that limit exists in this case? From the definition (Heine, Cauchy)? $\endgroup$ – guram Dec 16 '10 at 22:39
  • $\begingroup$ @guram: I believe you can prove that geometrically, the same way we do for $\sin x / x$. Of course, the geometric proof gives the limit too. Let me try to update the answer with that. $\endgroup$ – Aryabhata Dec 16 '10 at 22:48
  • $\begingroup$ @guram: I have updated the answer with a geometric proof. $\endgroup$ – Aryabhata Dec 16 '10 at 23:09

The Maclaurin series for $\tan x$ begins with $x$, and there's no $x^2$ term since it's an odd function. Thus $\tan x = x + O(x^3)$, and therefore $(\tan x - x)/x^2 = O(x) \to 0$ as $x\to 0$.

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    $\begingroup$ This is very nice, although under the hood it’s essentially the same as the L’Hôpital’s rule proof… $\endgroup$ – Peter LeFanu Lumsdaine Dec 16 '10 at 22:19
  • $\begingroup$ Rigorously demonstrating the MacLaurin series is much harder. $\endgroup$ – user1119 Dec 16 '10 at 22:32
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    $\begingroup$ @Peter: Hush, don't tell anyone...! ;) $\endgroup$ – Hans Lundmark Dec 17 '10 at 7:13
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    $\begingroup$ Another remark: L’Hôpital’s rule is a trivial consequence of Taylor's theorem in the case where both numerator and denominator are nice enough to have Taylor expansions. For problems like this one (and other "concrete" limits likely to be encountered in a calculus course) it is usually easier to just use the expansion. The strength of L’Hôpital's rule is that it holds under weaker assumptions, and also covers the case where the denominator tends to infinity instead of 0. $\endgroup$ – Hans Lundmark Dec 17 '10 at 7:27
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    $\begingroup$ @George: You don't need the MacLaurin series; all you need is $\sin x = x+O(x^3)$ and $\cos x = 1+O(x^2)$, from which you easily obtain $\tan x = x + O(x^3)$. This approach is a lot simpler than de l'Hospital since you don't have to compute any derivatives. $\endgroup$ – Hendrik Vogt Dec 18 '10 at 12:27

Note that ${\displaystyle {\tan(x) - x \over x^2}}$ is an odd function, so it suffices to show the limit from either side is zero. So we focus on the right limit, and changing $x$ to $\sqrt{x}$ it suffices to show that $$\lim_{x \rightarrow 0^+} {\tan(\sqrt{x}) - \sqrt{x} \over x} = 0$$ We apply the mean-value theorem to $f(y) = \tan(\sqrt{y}) - \sqrt{y}$ on $[0,x]$, which we can do since the mean-value theorem only requires that $f(x)$ is differentiable on the interior of the interval. We obtain that there is a $y \in (0,x)$ such that $$f'(y) = {\tan(\sqrt{x}) - \sqrt{x} \over x}$$ But using the chain rule we have $$f'(y) = {\sec^2(\sqrt{y}) - 1 \over 2 \sqrt{y}}$$ $$= {\tan^2(\sqrt{y}) \over \sqrt{y}}$$ Note that we have $$\lim_{y \rightarrow 0^+} {\tan^2(\sqrt{y}) \over \sqrt{y}} = \lim_{y \rightarrow 0^+}\tan(\sqrt{y})\,\,\,\times \,\,\lim_{y \rightarrow 0^+}{\tan(\sqrt{y}) \over \sqrt{y}}$$ $$ = 0*1 = 0$$ Thus we conclude that ${\displaystyle \lim_{x \rightarrow 0^+} {\tan(\sqrt{x}) - \sqrt{x} \over x} = 0}$ as needed.


Let's solve it elementarily considering $\sin(x)<x<\tan(x),\space 0< x <\frac{\pi}{2}.$ From this inequality we
get that: $$0\leq\lim_{x\to 0}{\frac{\tan{(x)}-x}{x^2}}\leq\lim_{x\to 0}{\frac{\tan{(x)}-\sin(x)}{x^2}} \tag1$$ $$\lim_{x\to 0}{\frac{\tan{(x)}-\sin(x)}{x^2}}=\lim_{x\to 0}{\frac{\sin(x)(\frac{1}{\cos(x)}-1)}{x^2}}=\lim_{x\to 0} \frac{\sin(x)}{x} \lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)}=$$ $$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)}=\lim_{x\to 0} \frac{1-\cos^2(x)}{x\cos(x)(1+\cos(x))}=\lim_{x\to 0} \frac{x\sin^2(x)}{x^2\cos(x)(1+\cos(x))}=0.\tag2$$

From $(1)$ and $(2)$ by applying Squeeze's theorem we get that $L=0$.

The proof is complete.

  • $\begingroup$ I see every part of your answer now except for the actual justification of replacing $x$ by $\sin x$ and claiming equality as opposed to some sort of bounding. It does tend to $x$, and it takes only one line with $\sin x = x + O(x^3)$. Perhaps this is an obvious step to you, meriting sweeping under the rug? $\endgroup$ – davidlowryduda Jun 12 '12 at 13:08
  • $\begingroup$ Yes, I do not take that line for granted. But it is true. Thank you for explaining things to me. +1 $\endgroup$ – davidlowryduda Jun 12 '12 at 13:19

I am not sure what you can use; if you know of Maclaurin series, the argument is very easy.

I believe that what your tutor meant is that you cannot rewrite $\displaystyle \frac{\tan x-x}{x^2}=\frac{\frac{\sin x}x-\cos x}{x\cos x}$ as $\displaystyle \frac{1-\cos x}{x\cos x}$. The argument below uses that $\sin x/x\to 1$, but not by substituting it this way.

Note that $\displaystyle \frac{1-\cos x}{x^2}=\frac{1-\cos^2x}{x^2(1+\cos x)}=\left(\frac{\sin x}x\right)^2\frac1{1+\cos x}\to\frac12$.

It is enough to find the limit of $\displaystyle \frac{\frac{\sin x}x-\cos x}x$. Using the limit above, all you need is to find the limit of $\displaystyle\frac{\sin x -x}{x^2}$, because $$\frac{\frac{\sin x}x-\cos x}x=\frac{\frac{\sin x}x+\left(\frac{1-\cos x}{x^2}\right)x^2-1}x.$$

The limit of $\displaystyle\frac{\sin x -x}{x^2}$ is 0. This is fairly easy to evaluate using the Maclaurin expansion of $\sin x$. All you really need is that $\sin x=x+O(x^3)$, but I am not sure you are familiar with this notation. Also, you could define $f(x)=\sin x/x$ if $x\ne 0$ and $f(0)=1$, and verify that this function is differentiable at 0. But I am not certain you have the tools to do that.


My inclination is to rewrite the original limit as:$$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}=\lim_{x\to 0}{\frac{\frac{\tan{x}}{x}-1}{x-0}},$$ with the thinking that if $f(x)=\frac{\tan x}{x}$, $\lim_{x\to 0}f(x)=1$ and $\lim_{x\to 0}f'(x)=0$ and $$f'(a)=\lim_{x\to a}{\frac{f(x)-f(a)}{x-a}}=\lim_{x\to a}{\frac{\frac{\tan x}{x}-\frac{\tan a}{a}}{x-a}}.$$ Unfortunately, to proceed from there, I'm getting stuck in some ugliness with double limits.


$lim_{x \to 0} \frac{tan(x)−x}{x*x}$ (0/0 Form)

$lim_{x \to 0} \frac{sec(x) * sec(x) -1}{2x}$ (0/0) Form $\frac{d(Numerator)}{d(Denominator)}$

$lim_{x \to 0} \frac{2secx * sec(x) *tan(x)}{2} = sec0 * sec0 *tan0 =1 * 1 * 0 =0$

Ans is 0.

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    $\begingroup$ Without explanation or math formatting this is very hard to read. Perhaps comparing with the older Answers would give some encouragement to learn MathJax and MarkDown techniques. $\endgroup$ – hardmath Apr 14 '14 at 12:28

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