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

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

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

Solution 1:

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

Solution 2:

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 $1$.

Solution 3:

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

Solution 4:

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

Solution 5:

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