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.
 A: $$\tan { x } =x+\frac { { x }^{ 3 } }{ 3 } +\frac { 2{ x }^{ 5 } }{ 15 } +\cdots \\ \frac { \tan { x }  }{ x } =\frac { x+\frac { { x }^{ 3 } }{ 3 } +\frac { 2{ x }^{ 5 } }{ 15 } +\cdots  }{ x } =1+\frac { { x }^{ 2 } }{ 3 } +\frac { 2{ x }^{ 4 } }{ 15 } +\cdots \\ \lim _{ x\rightarrow 0 }{ \left( \frac { \tan { x }  }{ x }  \right)  } =1$$Or for the geometric proof see:http://www.proofwiki.org/wiki/Limit_of_Sine_of_X_over_X/Geometric_Proof
A: 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 $\lim \limits_{x\to 0}\cos x=1$.  
A: 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 $$\lim _{ x\rightarrow 0 }{ \frac { \tan { x }  }{ x }  } =1.$$
A: Strong hint: $$\displaystyle \lim \limits_{x\to 0}\left(\frac{\tan (x)}{x}\right)=\lim \limits_{x\to 0}\left(\frac{\tan (x)-0}{x-0}\right)=\lim \limits_{x\to 0}\left(\frac{\tan(x)-\tan(0)}{x-0}\right)=\cdots$$
A: You could expand $tan(x)$ as a power series and then divide all terms by $x$ and then take the limit?
A: Here's mine!
$$\lim_{x \to 0} \frac{\tan x}{x}$$
$$= \lim_{x \to 0} \sec x\frac{\sin x}{x}$$
$$= \bigg(\lim_{x \to 0}\sec x\bigg) * \bigg(\lim_{x \to 0} \frac{\sin x}{x}\bigg)$$
$$= \bigg(\lim_{x \to 0}\cos x\bigg)^{-1} * 1 $$
$$= 1^{-1} * 1$$
$$= 1$$ :)
A: 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 $1$.

A: 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:
$$
\begin{align}
\lim_{x\to0}\frac{\tan(x)}{x}
&=\lim_{x\to0}\frac1{\cos(x)}\frac{\sin(x)}{x}\\
&=\frac1{\cos(0)}\lim_{x\to0}\frac{\sin(x)}{x}\\[6pt]
&=1
\end{align}
$$
A: 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.
A: Using L'Hôpital rule, we'll see
$$
\lim_{x\to\infty}\frac{\mathrm{tan}(x)}{x}=\lim_{x\to\infty}\frac{\left (\mathrm{cos}^{2}(x)  \right )^{-1}}{1}=\lim_{x\to\infty} 1-\mathrm{sin}^{2}(x)=\dots
$$
A: $$\lim_{x\rightarrow 0} \frac{\tan (x)}{x} = \frac{d}{dx}\tan(0)=\sec^2(0)=1$$
This, I find, is the simplest method of showing it...
A: 
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 $$ \lim_{n → \infty } \frac{\tan \left(\frac{π}{n}\right)}{\frac{π}{n}}=\lim_{x → 0 } \frac{\tan \left(x\right)}{x}=1$$
