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. \(\displaystyle \lim_{x \to 0}\frac{\tan x}{x}=1\)