# CBSE Class 11 - Mathematics - Limits and Derivatives Part-4- Limits of Trigonometric Functions

In the previous blog post Limits and derivatives Part-3, we learn about the algebra of limits. Now, let us study few theorems before we discuss about trignometric limits.

Theorem 1: Let f(x) and g(x) be two real valued functions with same domain such that f(x) ≤ g(x) for all x.
For some a, if $\lim_{x\rightarrow a} f(x)$ and $\lim_{x\rightarrow a} g(x)$ exist then $\lim_{x\rightarrow a} f(x) \leq \lim_{x\rightarrow a} g(x)$

Theorem 2: Sandwich Theorem or Squeeze Theorem
Let f(x) and g(x) be two real valued functions with same domain such that f(x) ≤ g(x) ≤ h(x) for all x in common domain.
For some a, if $\lim_{x\rightarrow a} f(x) = l = \lim_{x\rightarrow a} h(x)$, then $\lim_{x\rightarrow a} g(x) = l$

Theorem 3:$\lim_{x\rightarrow 0} sin x = 0$ and $\lim_{x\rightarrow 0} cos x = 1$ where x is measured in radians.

Theorem 4:  $\lim_{x\rightarrow 0} \frac{sin x}{x} = 1$
and
$\lim_{x\rightarrow 0} \frac{tan x}{x} = 1$

Theorem 5: $\lim_{x\rightarrow 0} \frac{1 - cos x}{x} = 0$

Let us solve some problems.

Q1: Evaluate $\lim_{x\rightarrow 0} \frac{sin 3x}{x}$

=$\lim_{x\rightarrow 0} \left ( 3 \times \frac{sin 3x}{3x} \right )$

=$3\lim_{x\rightarrow 0} \left ( \frac{sin 3x}{3x} \right )$

= 3(1)

= 3

Q2: Evaluate $\lim_{x\rightarrow 0} \left ( \frac{\sin ax}{ \sin bx} \right )$

=$\lim_{x\rightarrow 0} \left ( \frac{\sin ax}{ \sin bx} \right )$

= $\lim_{x\rightarrow 0} \left ( \frac{(\frac{\sin ax}{ax})ax}{ (\frac{\sin bx}{bx})bx} \right )$

= $\frac{\lim_{x\rightarrow 0} (\frac{\sin ax}{ax})ax}{ \lim_{x\rightarrow 0} (\frac{\sin bx}{bx})bx}$

= $\frac{a(1)}{b(1)}$
= $\frac{a}{b}$

Q3: Evaluate $\lim_{x\rightarrow 0} \frac{\sin 5x}{\tan 3x}$

=$\lim_{x\rightarrow 0} \frac{\sin 5x}{\tan 3x}$
= $\lim_{x\rightarrow 0} \frac{\frac{\sin 5x}{x}}{\frac{\tan 3x}{3x}}$
= $\frac{\lim_{x\rightarrow 0} \frac{\sin 5x}{x}}{\lim_{x\rightarrow 0}\frac{\tan 3x}{3x}}$
=$\frac{5}{3}.\frac{1}{1}$
=$\frac{5}{3}$

Q4: Evaluate $\lim_{x\rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x}$

Answer: Let $(x - \frac{\pi}{2}) = y$ so that when $x \rightarrow \frac{\pi}{2}$ then $y \rightarrow 0$

$\therefore \lim_{x\rightarrow 0} \frac{\cos (\frac{\pi}{2} + y)}{- y}$

= $\lim_{x\rightarrow 0} (\frac{-\sin y}{-y})$
= $\lim_{x\rightarrow 0} (\frac{\sin y}{y})$
= 1

Hope it helps you get a fair idea about limits of Trigonometric Functions. In the next post, we shall discuss about other problems
related to limits.