Limits of Trigonometric Functions

In some cases, solving the limits of trigonometric functions is almost the same as solving the limits of algebraic functions, for example by the direct substitution method or by the factoring method. Trigonometric formulas and limit theorems can help to solve the limits of trigonometric functions.

Direct Substitution Method

Example

Calculate the limit values of the following trigonometric functions.

1. $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \sin x$
2. $\mathop {\lim }\limits_{x \to 0} \left( {\cos 2x - 1} \right)$
3. $\mathop {\lim }\limits_{x \to 0} \left( {{{\sin }^2}x - {{\cos }^2}x} \right)$

Answer

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \sin x = \sin \frac{\pi }{2} = 1$

So, $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \sin x = 1$

$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \left( {\cos 2x - 1} \right) &= \cos 2.0 - 1\\ &= \cos 0 - 1\\ &= 1 - 1\\ &= 0 \end{array}$

So, $\mathop {\lim }\limits_{x \to 0} \left( {\cos 2x - 1} \right) = 0$

$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \left( {{{\sin }^2}x - {{\cos }^2}x} \right) &= {\sin ^2}0 - {\cos ^2}0\\ &= {0^2} - {1^2}\\ &= - 1 \end{array}$

So, $\mathop {\lim }\limits_{x \to 0} \left( {{{\sin }^2}x - {{\cos }^2}x} \right) = - 1 $

Factoring Method

Example

Calculate the limit values of the following trigonometric functions.

1. $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{\sin x}}$
2. $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{\sin x}}$

Answer

Using direct substitution method, we get:

• $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{\sin x}} = \frac{{\sin 2.0}}{{\sin 0} } = \frac{0}{0}$
• $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{\sin x}} = \frac{{1 - \cos 2.0}}{{ \sin 0}} = \frac{{1 - 1}}{0} = \frac{0}{0}$

By direct substitution we get the value $\frac{0}{0}$, which is of indefinite form. Therefore, it is necessary to make other efforts, namely as follows:

$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{\sin x}} &= \mathop {\lim }\limits_{x \to 0} \frac{{ 2\cancel{{\sin x}}\cos x}}{{\cancel{{\sin x}}}}\\ &= \mathop {\lim }\limits_{x \to 0} 2\cos x\\ &= 2\cos 0\\ &= 2.1\\ &= 2 \end{array}$

So, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{\sin x}} = 2$.

$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{\sin x}} &= \mathop {\lim }\limits_{x \to 0} \frac {{1 - \left( {1 - 2{{\sin }^2}x} \right)}}{{\sin x}}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}}{{\sin x}}\\ &= \mathop {\lim }\limits_{x \to 0} 2\sin x\\ &= 2\sin 0\\ &= 2.0\\ &= 0 \end{array}$

So $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{\sin x}} = 0$.

Limit Formulas for Trigonometric Functions

Limit trigonometric functions can also be solved using formulas. The limit formulas of the trigonometric functions in question are as follows:

\[\boxed{\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}} = 1\\ \mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan x}} = 1 \end{array}}\]

The limit formulas for trigonometric functions above can be extended. For example $u$ is a function of $x$ and if $x \to 0$ then $u \to 0$, the formulas can be written as follows:

\[\boxed{\begin{array}{l} \mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}{u} = \mathop {\lim }\limits_{u \to 0} \frac{u}{{\sin u}} = 1\\ \mathop {\lim }\limits_{u \to 0} \frac{{\tan u}}{u} = \mathop {\lim }\limits_{u \to 0} \frac{u}{{\tan u}} = 1 \end{array}}\]

Example

Calculate the limit values ​​of the following trigonometric functions.

1. $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{2x}}$
2. $\mathop {\lim }\limits_{x \to 0} \frac{{6x}}{{x + \sin 3x}}$

Answer

1. For example $u = 4x \to x = \frac{1}{4}u$. If $x \to 0$ then $u \to 0$, so
$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{2x}} &= \mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}{{2\left( {\frac{1}{4}u} \right)}}\\ &= \mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}{{\frac{1}{2}u}}\\ &= \mathop {\lim }\limits_{u \to 0} 2\frac{{\sin u}}{u}\\ &= 2\mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}{u}\\ &= 2.1\\ &= 2 \end{array}$

So, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{2x}} = 2$.

$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{6x}}{{x + \sin 3x}} &= \mathop {\lim }\limits_{x \to 0} \frac{1 }{{\frac{{x + \sin 3x}}{{6x}}}}\\ &= \frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \frac{{x + \sin 3x}} {{6x}}}}\\ &= \frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{ 6x}} + \frac{{\sin 3x}}{{6x}}} \right)}}\\ &= \frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \frac{1}{6} + \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{2.3x}}}}\\ &= \frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \frac{1}{6} + \frac {1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}}}\\ &= \frac{1}{{\frac{1}{6} + \frac{1}{2}.1}}\\ &= \frac{1}{{\frac{4}{6}}}\\ &= \frac{6}{4}\\ &= 1\frac{1}{2} \end{array}$

So, $\mathop {\lim }\limits_{x \to 0} \frac{{6x}}{{x + \sin 3x}} = 1\frac{1}{2}$.