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

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 $1$.

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 $\lim \limits_{x\to 0}\cos x=1$.

$$\tan { x } =x+\frac { { x }^{ 3 } }{ 3 } +\frac { 2{ x }^{ 5 } }{ 15 } +\cdots \\ \frac { \tan { x } }{ x } =\frac { x+\frac { { x }^{ 3 } }{ 3 } +\frac { 2{ x }^{ 5 } }{ 15 } +\cdots }{ x } =1+\frac { { x }^{ 2 } }{ 3 } +\frac { 2{ x }^{ 4 } }{ 15 } +\cdots \\ \lim _{ x\rightarrow 0 }{ \left( \frac { \tan { x } }{ x } \right) } =1$$Or for the geometric proof see:http://www.proofwiki.org/wiki/Limit_of_Sine_of_X_over_X/Geometric_Proof

Consider the following circle with a regular $n$ side polygon inside:

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 $$\lim _{ x\rightarrow 0 }{ \frac { \tan { x } }{ x } } =1.$$

You could expand $tan(x)$ as a power series and then divide all terms by $x$ and then take the limit?

Here’s mine!

$$\lim_{x \to 0} \frac{\tan x}{x}$$

$$= \lim_{x \to 0} \sec x\frac{\sin x}{x}$$

$$= \bigg(\lim_{x \to 0}\sec x\bigg) * \bigg(\lim_{x \to 0} \frac{\sin x}{x}\bigg)$$

$$= \bigg(\lim_{x \to 0}\cos x\bigg)^{-1} * 1 $$

$$= 1^{-1} * 1$$

$$= 1$$ 🙂

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}

$$

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.

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

$$

$$\lim_{x\rightarrow 0} \frac{\tan (x)}{x} = \frac{d}{dx}\tan(0)=\sec^2(0)=1$$

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

We know that if the polygon have more sides the its perimeter will get closer to the perimeter of circle.

$$\frac{HB}{R}=\tan(\frac{360}{2n})=\tan(\frac{\pi}{n})\\AB=2HB=2R\tan(\frac{\pi}{n})$$

Perimeter of polygon = $nAB=2nR\tan(\frac{\pi}{n}) \\$

Perimeter of circle = $2\pi R $

Now when $n \to \infty $ $$ \lim_{n \to \infty } \frac{\text{Perimeter of

polygon}}{\text{Perimeter of circle}}=1$$

$$\lim_{n \to \infty } \frac{2n R \tan(\frac{\pi}{n})}{2 \pi R }=1$$

$$\lim_{n \to \infty } \frac{n \tan(\frac{\pi}{n})}{ \pi }=1$$

$$\lim_{n \to \infty } \frac{\tan(\frac{\pi}{n})}{\frac{\pi}{n}}=1$$

Obviously $\frac{\pi}{n} \rightarrow 0$ name as $x$ so $$ \lim_{n \to \infty } \frac{\tan(\frac{\pi}{n})}{\frac{\pi}{n}}=\lim_{x \to 0 } \frac{\tan(x)}{x}=1$$

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