Use the direct substation approach to find the limit of the sine of angle five times $x$ minus sine of angle three times $x$ divided by $x$ as the value of $x$ is closer to zero.

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

According to the direct substation method, it is evaluated that the limit of the quotient of the subtraction of sine of angle three times $x$ from the sine of angle five times $x$ by $x$ as the value of $x$ approaches to zero is indeterminate. It expresses that the direct substitution method is not useful to find the limit of the given rational function. So, it is better to think for an alternative mathematical approach.

The rational function can be split as the difference of two terms by distributing the expression in the denominator to both terms in the numerator.

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

The limiting operation can be distributed to the difference of the fractions to find the limit of the difference of the functions by calculating the difference of their limits as per the difference rule of the limits.

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

Look at the rational function in each term of the mathematical expression. The expression in every term closely matches with trigonometric limit rule in sine function. The angles inside the sine functions in the first and second terms are $5x$ and $3x$ respectively. Hence, the denominators in both terms should be $5x$ and $3x$ respectively.

Let’s make some acceptable changes for converting the rational functions into required form.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(1 \times \dfrac{\sin{5x}}{x}\Big)}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(1 \times \dfrac{\sin{3x}}{x}\Big)}$

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

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

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

The numbers $5$ and $3$ are coefficients of the rational functions, and they are constants. Therefore, they can be separated from the limit operations as per the constant multiple limit rule.

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

In the first term, the angle inside the sine function is $5x$ and the same angle $5x$ is also there in the denominator. In the second term, the angle inside the sine function is $3x$, and the same angle $3x$ is also there in the denominator. The two rational functions are similar to the trigonometric limit rule in sine function.

However, the trigonometric limit rule cannot be applied to both terms because the input of the limit operation should also be equal to either the angle inside the sine function or the function in the denominator. Therefore, let’s try to make some adjustments to apply it to each term.

If $x \,\to\, 0$ then $5 \times x \,\to\, 5 \times 0$. Therefore, $5x \,\to\, 0$. Similarly, if $x \,\to\, 0$ then $3 \times x \,\to\, 3 \times 0$. Therefore, $3x \,\to\, 0$. It clears that when the value of $x$ tends to $0$, the values of both $5x$ and $3x$ also approach to $0$. Therefore, the inputs of the limit operations can be set to $5x$ and $3x$ respectively instead of $x$.

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

Now, use the trigonometric limit rule in sine function to find the limit of each function. The limit of sine of angle $5x$ by $5x$ as the value of $5x$ approaches to $0$ is equal to $1$. Similarly, the limit of sine of angle $3x$ divided by $3x$ as the value of $3x$ is closer to $0$ is equal to $1$.

$=\,\,\,$ $5 \times 1$ $-$ $3 \times 1$

Now, simplify the arithmetic expression to find the limit of the given trigonometric rational function as the value of $x$ approaches to $0$.

$=\,\,\,$ $5-3$

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

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