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How to prove that $$\lim\limits_{x\to0}\frac{\tan x}x=1?$$

I'm looking for a method besides L'Hospital's rule.

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    $\begingroup$ How do you prove $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$? $\endgroup$ Commented Jul 20, 2013 at 17:37
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    $\begingroup$ Rewrite as $\frac{1}{\cos x}\frac{\sin x}{x}$. (I am assuming here that you have dealt in class with $\lim_{x\to 0}\frac{\sin x}{x}$.) $\endgroup$ Commented Jul 20, 2013 at 17:38
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    $\begingroup$ Power series will also work. $\endgroup$
    – Potato
    Commented Jul 20, 2013 at 17:40
  • $\begingroup$ @Potato No! I'm not allowed to use power series too $\endgroup$ Commented Jul 20, 2013 at 17:42
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    $\begingroup$ What is your definition of $\tan$? This will affect the answer to the question... $\endgroup$
    – GEdgar
    Commented Jul 20, 2013 at 20:45

12 Answers 12

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Strong hint: $$\displaystyle \lim \limits_{x\to 0}\left(\frac{\tan (x)}{x}\right)=\lim \limits_{x\to 0}\left(\frac{\tan (x)-0}{x-0}\right)=\lim \limits_{x\to 0}\left(\frac{\tan(x)-\tan(0)}{x-0}\right)=\cdots$$

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    $\begingroup$ Nice alt method, +1! $\endgroup$ Commented Jul 20, 2013 at 17:45
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    $\begingroup$ that's creative (: $\endgroup$
    – sigmatau
    Commented Jul 20, 2013 at 17:54
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    $\begingroup$ that's creative:) $\endgroup$
    – Alex
    Commented Jul 20, 2013 at 18:08
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    $\begingroup$ Hmm. How do you find the derivative of $\tan x$ evaluated at $x=0$ without knowing the value of $\lim_{x\rightarrow0}{\sin x\over x}$? $\endgroup$ Commented Jul 20, 2013 at 18:25
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    $\begingroup$ @Alex That's what I mean. I think you missed Mitra's point, though. He's implying that in order to know that the derivative of $\sin$ is $\cos$, etc, you need to know before hand that $\lim \limits_{x\to 0}\frac{\sin (x)}{x}=1$. $\endgroup$
    – Git Gud
    Commented Jul 20, 2013 at 18:28
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Consider the unit circle with center $O$. Let $A$ be a fixed point on the circumference. Let $X$ be a point on the circumference such that $\angle AOX = x$.

Let the tangent at $X$ intersect $OA$ extended at $B$. Since $\angle OXB = 90^\circ$ hence $BX = \tan x$.

Then, the area of the sector $OAX$ is $\frac{x\times 1^2}{2}$ and the area of the triangle $OXB$ is $\frac{1 \times \tan x}{2}$. It is clear that as $X$ tends towards $A$, the limit of these areas is $1$.

enter image description here

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  • $\begingroup$ GeoGebra, right? $\endgroup$ Commented Jul 20, 2013 at 19:15
  • $\begingroup$ @SohamChowdhury Yup :) I'm learning to use it; this is a screenshot of it. Apparrantly there is a way to share a web version which allows others to play around with the figures you draw. $\endgroup$
    – Calvin Lin
    Commented Jul 20, 2013 at 23:08
  • $\begingroup$ +1 I voted for this one Calvin because for the student who are facing this limit, we should apply geometric approach. Thanks for sharing us this way. $\endgroup$
    – Mikasa
    Commented Sep 8, 2013 at 7:20
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$$\tan { x } =x+\frac { { x }^{ 3 } }{ 3 } +\frac { 2{ x }^{ 5 } }{ 15 } +\cdots \\ \frac { \tan { x } }{ x } =\frac { x+\frac { { x }^{ 3 } }{ 3 } +\frac { 2{ x }^{ 5 } }{ 15 } +\cdots }{ x } =1+\frac { { x }^{ 2 } }{ 3 } +\frac { 2{ x }^{ 4 } }{ 15 } +\cdots \\ \lim _{ x\rightarrow 0 }{ \left( \frac { \tan { x } }{ x } \right) } =1$$Or for the geometric proof see:http://www.proofwiki.org/wiki/Limit_of_Sine_of_X_over_X/Geometric_Proof

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  • $\begingroup$ Simplest proof, +1 $\endgroup$
    – Phaptitude
    Commented Nov 10, 2013 at 23:32
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In order to find the derivative of $\sin x$, many calculus courses start by proving, sort of, that $$\lim \limits_{x\to 0}\frac{\sin x}{x}=1.\tag{1}$$

If that is already taken as "known" in your course, note that unless $\cos x=0$, we have
$$\frac{\tan x}{x}=\frac{1}{\cos x}\frac{\sin x}{x}.$$ Now we can take the limit. Use (1) and the fact that $\cos x$ is continuous at $0$ and therefore $\lim \limits_{x\to 0}\cos x=1$.

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  • $\begingroup$ You were a tiny bit faster! $\endgroup$ Commented Jul 20, 2013 at 17:50
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    $\begingroup$ @SohamChowdhury Check the comments to the question. $\endgroup$
    – Git Gud
    Commented Jul 20, 2013 at 17:51
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    $\begingroup$ @GitGud: Thanks for improving the TeX. $\endgroup$ Commented Jul 20, 2013 at 18:15
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Consider the following circle with a regular $n$ side polygon inside: enter image description here

We know that if the polygon have more sides the its perimeter will get closer to the perimeter of circle. $$\lim _{ n\rightarrow \infty }{ \frac { Perimeter\ of\ polygon }{ Perimeter\ of\ circle } } =\lim _{ n\rightarrow \infty }{ \frac { 2n\sin { \frac { \pi }{ n } } }{ 2\pi } } =\lim _{ n\rightarrow \infty }{ \frac { \sin { \frac { \pi }{ n } } }{ \frac { \pi }{ n } } }=1. $$ Assume $x=\frac { \pi }{ n } $ then we get $$\lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } } =1.$$We already know $\lim _{ x\rightarrow 0 }{ \cos x } =1$, therefore $$\lim _{ x\rightarrow 0 }{ \frac { \tan { x } }{ x } } =1.$$

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    $\begingroup$ If you use a polygon external to the circle you get $\tan x$ directly. $\endgroup$ Commented Jul 20, 2013 at 21:12
  • $\begingroup$ @MarkBennet Ah yes, I didn't think. Thank you. $\endgroup$
    – newzad
    Commented Jul 20, 2013 at 21:14
  • $\begingroup$ It is not correct to write "say $x=\pi n^{-1}$". $\endgroup$
    – Pedro
    Commented Jul 20, 2013 at 21:44
  • $\begingroup$ @PeterTamaroff the word "Say" is not correct? $\endgroup$
    – newzad
    Commented Jul 20, 2013 at 21:46
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    $\begingroup$ @newzad Ah, sorry, of course, but I think you know what I meant. The statement $f(x)\to \ell \;\;;x\to a$ is equivalent to $f(a_n)\to \ell$ for every $a_n\to a$; $a_n\neq a$. $\endgroup$
    – Pedro
    Commented Jul 20, 2013 at 21:54
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You could expand $tan(x)$ as a power series and then divide all terms by $x$ and then take the limit?

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    $\begingroup$ Sorry, just seen that you're not allowed to use the power series. $\endgroup$ Commented Jul 20, 2013 at 17:43
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Here's mine!

$$\lim_{x \to 0} \frac{\tan x}{x}$$ $$= \lim_{x \to 0} \sec x\frac{\sin x}{x}$$ $$= \bigg(\lim_{x \to 0}\sec x\bigg) * \bigg(\lim_{x \to 0} \frac{\sin x}{x}\bigg)$$ $$= \bigg(\lim_{x \to 0}\cos x\bigg)^{-1} * 1 $$ $$= 1^{-1} * 1$$ $$= 1$$ :)

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This limit is proven in this answer. That answer was in response to the question of how to show $$ \lim_{x\to0}\frac{\sin(x)}{x}=1 $$ However, since $\cos(x)$ is continuous at $x=0$, the two questions are related: $$ \begin{align} \lim_{x\to0}\frac{\tan(x)}{x} &=\lim_{x\to0}\frac1{\cos(x)}\frac{\sin(x)}{x}\\ &=\frac1{\cos(0)}\lim_{x\to0}\frac{\sin(x)}{x}\\[6pt] &=1 \end{align} $$

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One way to look at it is to consider an angle subtended by two finite lines, both of magnitude r, where the angle between them is x (we take x to be small). If you draw this out, you can see there are "3 areas" you can consider. One is the area enclosed with a straight line joining the two end points, an arc and lastly considering a right-angled triangle. Sorry I cant provide a diagram, I'm new to maths.stackexchange :)

you get the following result

1/2*r^2sinx < 1/2*r^2x < 1/2*r^2tanx for small x, with simplication we get

sinx < x < tanx divide by tanx yeilds

cosx < x/tanx < 1 taking the limit as x goes to 0, (which we can do as we took x to be small)

we get 1 < x/tanx < 1, by squeeze theorem this tells us the limit of as x >>0 for x/tanx is 1. Now the limit of tans/x as x approaches 0 will be the reciprocal of this. I should mention I am assuming early foundational results regarding limits in an Analysis course. Hence, the limit is 1.

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Using L'Hôpital rule, we'll see $$ \lim_{x\to\infty}\frac{\mathrm{tan}(x)}{x}=\lim_{x\to\infty}\frac{\left (\mathrm{cos}^{2}(x) \right )^{-1}}{1}=\lim_{x\to\infty} 1-\mathrm{sin}^{2}(x)=\dots $$

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    $\begingroup$ The question specifies "other than L'hopital" $\endgroup$
    – robjohn
    Commented Jul 20, 2013 at 18:30
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$$\lim_{x\rightarrow 0} \frac{\tan (x)}{x} = \frac{d}{dx}\tan(0)=\sec^2(0)=1$$

This, I find, is the simplest method of showing it...

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  • $\begingroup$ Possible duplicate of this. $\endgroup$
    – Git Gud
    Commented Jul 20, 2013 at 18:37
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enter image description here

We know that if the polygon have more sides the its perimeter will get closer to the perimeter of circle. $$\frac{HB}{R}=\tan \left(\frac{360}{2n}\right)=\tan \left(\frac{π}{n}\right)\\AB=2HB=2R\tan \left(\frac{π}{n}\right)$$ Perimeter of polygon = $nAB=2nR\tan \left(\frac{π}{n}\right) \\$

Perimeter of circle = $2π R $

Now when $n → \infty $ $$ \lim_{n \to \infty } \frac{\text{Perimeter of polygon}}{\text{Perimeter of circle}}=1$$ $$\lim_{n → \infty } \frac{2n R \tan \left(\frac{π}{n}\right)}{2 π R }=1$$ $$\lim_{n → \infty } \frac{n \tan \left(\frac{π}{n}\right)}{ π }=1$$ $$\lim_{n → \infty } \frac{\tan \left(\frac{π}{n}\right)}{\frac{π}{n}}=1$$

Obviously $\frac{π}{n} → 0$ name as $x$ so $$ \lim_{n → \infty } \frac{\tan \left(\frac{π}{n}\right)}{\frac{π}{n}}=\lim_{x → 0 } \frac{\tan \left(x\right)}{x}=1$$

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  • $\begingroup$ Nice geometrical solution. Thanks $\endgroup$ Commented Jul 24, 2015 at 8:30
  • $\begingroup$ thanks mahdi , are you in sharif ? $\endgroup$
    – Khosrotash
    Commented Jul 24, 2015 at 8:40
  • $\begingroup$ what is down vote for ? (I don't think it's duplicated ) $\endgroup$
    – Khosrotash
    Commented Jul 24, 2015 at 23:23
  • $\begingroup$ Such geometric proofs always seem artificial to me because you need to rigorously show the equivalence of the concepts/properties of geometry and calculus. In particular, what does it mean to take a polygon "in the limit" ? And how are the sine/cosine of geometry equivalent to those of calculus ? $\endgroup$
    – user65203
    Commented Feb 18, 2020 at 7:56

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