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

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

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 $1$.

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