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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*}

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    Just out of curiosity, how did you get to that limit?2010-12-16
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    In the third term of the last line, a cosine needs a square.2010-12-16
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    @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...2010-12-16
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    In the numerator you have two of those.2010-12-16
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    @Rasmus: Of course You are right, I fix that mistake. Thank You.2010-12-16
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    Very nice problem and very tricky if you plan to solve it in an easy way. It took me some time.2012-06-12

7 Answers 7

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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...

Consider

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

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

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

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

Thus

$$L = 0$$

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

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

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    @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.2010-12-16
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    @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...2010-12-16
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    @Moron: But how to show that limit exists in this case? From the definition (Heine, Cauchy)?2010-12-16
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    @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.2010-12-16
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    @guram: I have updated the answer with a geometric proof.2010-12-16
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    @Moron: Wow, that is great answer! Thank You very much, You realy helped me.2010-12-16
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    @guram: Thanks for the nice problem. I would have upvoted your question, but I am out of votes for today!2010-12-16
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    This was nice, thanks!2010-12-17
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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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    This is very nice, although under the hood it’s essentially the same as the L’Hôpital’s rule proof…2010-12-16
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    Rigorously demonstrating the MacLaurin series is much harder.2010-12-16
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    @Peter: Hush, don't tell anyone...! ;)2010-12-17
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    @George: Compare the proofs of L’Hôpital’s rule (Theorem 5.13) and Taylor's Theorem (Theorem 5.15) in Rudin's *Principles*. They are approximately equal in length, and I would say that the latter proof is actually simpler!2010-12-17
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    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 0.2010-12-17
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    @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.2010-12-18
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    @Hendrik Vogt: Is it possible to show that without Taylor's theorem?2010-12-18
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    @George: Could you please specify which part of my comment you mean? The one about sin/cos or the conclusion for tan?2010-12-18
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    @Hendrik Vogt: As an example, the case of $sin\ x$ would be enough. Let me be optimistic that the cases of cos and tan will be similar.2010-12-18
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    @George: OK, that depends a lot on what definition of sin you use. Being an analysis guy, I favour the definition $\sin x := \mathop{Im}e^{ix} = \sum_{n=0}^\infty (-1)^nx^{2n+1}/(2n+1)!$ for real $x$; then $\sin x = x+O(x^3)$ becomes almost trivial. What definition did you learn?2010-12-18
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    @Hendrik Vogt: I see that you are defining these functions using the infinite series. Then of course it is trivial. Moron's definition as the ratio of opposite side by hypotenuse is the one which is more geometric.2010-12-18
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    @George: It's just that the precise geometric _definition_ of sin and cos is a lot more difficult than the analytic one, as far as I know. Please tell me when I'm wrong.2010-12-18
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    @Hendrik Vogt: In the calculus course I had, the derivative of sin and cos were computed geometrically. One can use the sandwich theorem for the limit and the angle addition formula. If you use the analytic series definitions for these functions, then all these limit computations at 0 are much simpler, aren't they? The geometric approach glossing over Euclidean geometry subtleties might not be completely rigorous; but that seems to be the way adopted in many courses and textbooks.2010-12-18
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    @GeorgeS. You could do this: $\cos x \le 1$, now integrating from $0$ to $x$ we have $\sin x \le x$. Integrating again, $1-\cos x \le \frac {x^2} 2 \implies \cos x \ge 1 - \frac {x^2} 2$, and continuining on $\sin x \ge x - \frac {x^3} {3!}$2012-06-13
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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.

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Let's solve it elementarily considering $\sin(x) 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.

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    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?2012-06-12
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    Yes, I do not take that line for granted. But it is true. Thank you for explaining things to me. +12012-06-12
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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.

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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.

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$lim_{x \to 0} \frac{tan(x)−x}{x*x}$ (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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    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.2014-04-14