Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}-\sin{3x}}{x}}$ by the Trigonometric identities

Firstly, let us check the process of finding the limit of the quotient of sine of five times $x$ minus sine of three times $x$ by $x$ as the value of $x$ approaches to zero by the direct substitution method.

$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}-\sin{3x}}{x}}$ $\,=\,$ $\dfrac{0}{0}$

As per the direct substitution method, it is calculated that the limit is indeterminate for the sine of five times $x$ minus sine of angle three times $x$ divided by $x$ as the value of $x$ is closer to zero.

It expresses that calculating the limit for the given trigonometric function in rational function is not recommendable and it indicates us to think for alternative method.

Convert the difference into Product form

The difference of the sine functions in the numerator and expression in the denominator are the main reason for getting the limit of the given trigonometric rational function as indeterminate.

The difference of sine functions can be transformed into product form of the trigonometric functions as per the difference to product trigonometric identity in sine functions.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2\cos{\bigg(\dfrac{5x+3x}{2}\bigg)}\sin{\bigg(\dfrac{5x-3x}{2}\bigg)}}{x}}$

The expression in the denominator is a variable. So, no need to take any action on this.

Simplify the Trigonometric function

It is time to simplify the mathematical expression and it helps us to find the limit mathematically.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2\cos{\bigg(\dfrac{8x}{2}\bigg)}\sin{\bigg(\dfrac{2x}{2}\bigg)}}{x}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2\cos{\bigg(\dfrac{\cancel{8}x}{\cancel{2}}\bigg)}\sin{\bigg(\dfrac{\cancel{2}x}{\cancel{2}}\bigg)}}{x}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2\cos{(4x)}\sin{(x)}}{x}}$

Find the Limits of the functions

There is a sine function in numerator and the angle inside the sine function is also there in the denominator. They indicate us to separate them from the rational function. The separation plays a vital role in finding limit of the given trigonometric rational function.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2\cos{(4x)} \times \sin{(x)}}{x}}$

Now, the rational function can be split as a product of two functions as per the multiplication rule of the fractions.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(2\cos{(4x)} \times \dfrac{\sin{(x)}}{x}\bigg)}$

Use product rule of the limits to find the limit of the product of two functions by the product of their limits.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize 2\cos{(4x)}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{(x)}}{x}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize 2\cos{(4x)}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$

The mathematical expression consists of two factors. Each factor expresses the limit of a function. So, let’s find the limit of each function one by one. Firstly, let us find the limit of the function at the position of first factor and it can be done by the direct substitution.

$=\,\,\,$ $2\cos{\big(4(0)\big)}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$

$=\,\,\,$ $2\cos{(4 \times 0)}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$

$=\,\,\,$ $2\cos{(0)}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$

$=\,\,\,$ $2\cos{0}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$

$=\,\,\,$ $2 \times \cos{0}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$

According to the trigonometry, the cosine of zero radian is equal to one. So, substitute it in the mathematical expression.

$=\,\,\,$ $2 \times 1$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$

$=\,\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$

As per the trigonometric limit rule in sine function, the limit of the sine of angle $x$ divided by $x$ as the value of $x$ approaches zero is equal to one.

$=\,\,\,$ $2 \times 1$

$=\,\,\,$ $2$