# Limit of trigonometric function

Trigonometry is branch of mathematics that deals about Triangle. The trigonometric ratio with reference to an angle x is called trigonometric function. For example,
f(x)= sinx

In this section we learn about two very specific but important trigonometric limits, and how to use them; and other tricks to find most other limits of trigonometric functions. The first involves the sine function, and the limit is
$\displaystyle \lim_{\theta \to 0}\frac{\sin \theta }{\theta}=1$

Here's a graph of $f(x)=\frac{\sin x }{x}$, showing that it has a hole at x = 0. Our task in this section will be to prove that the limit from both sides of this function is 1.

#### Theorems on Limit of trigonometric function

Area if triangle $OPM=\frac{1}{2} \sin \theta \cos \theta$

त्रिभुजको क्षेत्रफल 1/2 * आधार * उचाई हुने भएकोले $\triangle OPM=\frac{1}{2} \sin \theta \cos \theta$ हुन्छ ।

Area if sector $OPB=\frac{1}{2} \theta$

वृतको चाँदक्षेत्रको क्षेत्रफल 1/2 * अर्धव्यास 2 * केन्द्रिय कोण हुने भएकोले sector $OPB=\frac{1}{2} \theta$ हुन्छ ।

Area if triangle $OBA=\frac{1}{2} \tan \theta$

त्रिभुजको क्षेत्रफल 1/2 * आधार * उचाई हुने भएकोले $\triangle OBA=\frac{1}{2} \tan \theta$ हुन्छ ।

In the figure above
Area of triangle OMP=$\frac{1}{2} \sin \theta \cos \theta$
Area of sector OAP=$\frac{1}{2} \theta$
Area of triangle OAB=$\frac{1}{2} \tan \theta$
Now
Area of triangle OMP $\le$ Area of sector OAP $\le$ Area of triangle OAB
or $\frac{1}{2} \sin \theta \cos \theta \le \frac{1}{2} \theta \le \frac{1}{2} \tan \theta$
or $\sin \theta \cos \theta \le \theta \le \tan \theta$
or $\cos \theta \le \frac{\theta}{\sin \theta } \le \frac{1}{\cos \theta}$
or $\frac{1}{\cos \theta} \ge \frac{\sin \theta }{\theta} \ge \cos \theta$
Taking limit as $\theta \to 0$, we get
$\displaystyle \lim_{\theta \to 0} \frac{1}{\cos \theta} \ge \lim_{\theta \to 0}\frac{\sin \theta }{\theta} \ge \lim_{\theta \to 0} \cos \theta$
or $\displaystyle \frac{1}{\cos 0} \ge \lim_{\theta \to 0}\frac{\sin \theta }{\theta} \ge \cos 0$
or $\displaystyle \frac{1}{1} \ge \lim_{\theta \to 0}\frac{\sin \theta }{\theta} \ge 1$
or $\displaystyle 1 \ge \lim_{\theta \to 0}\frac{\sin \theta }{\theta} \ge 1$
or $\displaystyle \lim_{\theta \to 0}\frac{\sin \theta }{\theta}=1$
This completes the proof.

#### More Theorems on Limit of trigonometric function

1. The another important limit involves the cosine function, specifically the function
$\displaystyle \lim_{\theta \to 0}\frac{\cos \theta -1}{\theta}=0$

Here's a graph of $f(x)=\frac{\cos x-1}{x}$, showing that it has a hole at x = 0. Our task in this section will be to prove that the limit from both sides of this function is 0.

Prove that
$\displaystyle \lim_{x \to 0}\frac{1- \cos x}{x}=0$
Solution
The limit is
$\displaystyle \lim_{x \to 0}\frac{1- \cos x}{x}$
or $\displaystyle \lim_{x \to 0}\frac{1- \cos x}{x} \times \frac{1+ \cos x}{1+ \cos x}$
or $\displaystyle \lim_{x \to 0}\frac{1- \cos^2 x}{x(1+ \cos x)}$
or $\displaystyle \lim_{x \to 0}\frac{\sin^2 x}{x(1+ \cos x)}$
or $\displaystyle \lim_{x \to 0}\frac{\sin x}{x} \times \lim_{x \to 0} \frac{\sin x}{1+ \cos x}$
or $1 \times \frac{0}{1+ 1}$
or 0
This completes the proof
2. $\displaystyle \lim_{x \to 0} \sin x=0$
3. $\displaystyle \lim_{x \to 0} \cos x=1$
4. $\displaystyle \lim_{x \to 0}\frac{\tan x}{x}=1$