The limit of sine of angle $5$ times $x$ minus sine of angle $3$ times $x$ divided by $x$ as the value of variable $x$ approaches to zero is indeterminate as per the limit by direct substitution method.
$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}-\sin{3x}}{x}}$ $\,=\,$ $\dfrac{0}{0}$
The indeterminate form is mainly caused by the difference between the sine functions in the numerator, and a variable in the denominator.
In numerator, the expression inside each sine function is a linear expression in one variable. Similarly, the expression in denominator is also a linear expression in one variable. The both expressions have same degree. So, if the rational expression is split in such way that each sine function with the expression in denominator as a fraction, then it is possible to find the limit of the given rational function.
The rational function can be split as the difference of two rational functions as per the difference rule of the like fractions.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{5x}}{x}-\dfrac{\sin{3x}}{x}\Big)}$
Now, each rational function in the expression matches with the rational function in the sine function based trigonometric limit rule. So, it is time to find the limit of each rational function.
The difference rule of the limits can be used to find the limit of difference between two functions by finding the difference between their limits.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}}{x}}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{3x}}{x}}$
The coefficient of $x$ is $5$ inside the sine function in the first term and the coefficient of $x$ is $3$ inside the sine function in the second term. In order to use the sine function based trigonometric limit rule, the coefficient of $x$ should also be there in denominator of each rational function in every term of the expression.
Let’s use a trick to make denominator same as the expression inside the sine function in every term of the expression. According to a property of factors, one is a factor of every quantity. So, each rational function in every term can be expressed in product form with one.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(1 \times \dfrac{\sin{5x}}{x}\Big)}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(1 \times \dfrac{\sin{3x}}{x}\Big)}$
The coefficient of $x$ should be $5$ in the denominator of the rational function in the first term and the coefficient of $x$ should be $3$ in the denominator of the rational function in the second term of the mathematical expression. So, the factor $1$ can be written as the quotient of $5$ divided by $5$ in the first term and the factor $1$ can be written as the quotient of $3$ divided by $3$ in the second term.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{5}{5} \times \dfrac{\sin{5x}}{x}\Big)}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{3}{3} \times \dfrac{\sin{3x}}{x}\Big)}$
$1$ and $5$ are factors of $5$. So, the number $5$ can be expressed in the form of their factors as per factorization. $1$ and $3$ are factors of $3$. So, the number $3$ can also be expressed in the form of their factors as per factorisation.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{5 \times 1}{1 \times 5} \times \dfrac{\sin{5x}}{x}\Big)}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{3 \times 1}{1 \times 3} \times \dfrac{\sin{3x}}{x}\Big)}$
Now, each fraction in arithmetic form can be split as product of two fractions as per the product rule of the fractions.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{5}{1} \times \dfrac{1}{5} \times \dfrac{\sin{5x}}{x}\Big)}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{3}{1} \times \dfrac{1}{3} \times \dfrac{\sin{3x}}{x}\Big)}$
The quotient of $5$ divided by $1$ is equal to $5$ and the quotient of $3$ divided by $1$ is equal to $3$.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(5 \times \dfrac{1}{5} \times \dfrac{\sin{5x}}{x}\Big)}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(3 \times \dfrac{1}{3} \times \dfrac{\sin{3x}}{x}\Big)}$
Now, multiply the fractions inside the limit operations by using the product rule of the fractions in each term of the expression to find their products.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(5 \times \dfrac{1 \times \sin{5x}}{5 \times x}\Big)}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(3 \times \dfrac{1 \times \sin{3x}}{3 \times x}\Big)}$
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(5 \times \dfrac{\sin{5x}}{5x}\Big)}$ $-$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(3 \times \dfrac{\sin{3x}}{3x}\Big)}$
The numbers $5$ and $3$ are coefficients of the rational functions, and they are constants.
The constants can be separated from the limit operations as per the constant multiple limit rule.
$=\,\,$ $\displaystyle 5 \times \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}}{5x}}$ $-$ $\displaystyle 3 \times \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{3x}}{3x}}$
The expression in the denominator is same as the expression inside the sine function in both terms of the mathematical expression. However, it is not possible for us to use the sine function based trigonometric limit rule because the input of the limit operation should also be in same form. So, try to set input value of each limit operation in the expression.
$(a).\,\,$ If $x \,\to\, 0$ then $5 \times x \,\to\, 5 \times 0$. Therefore, $5x \,\to\, 0$.
$(b).\,\,$ Similarly, if $x \,\to\, 0$ then $3 \times x \,\to\, 3 \times 0$. Therefore, $3x \,\to\, 0$.
The above two mathematical statements express that the value of $5x$ also approaches to zero when the value of $x$ approaches to zero. Similarly, the value of $3x$ also approaches to zero when the value of $x$ approaches to zero.
Now, the input value can be given as $5x$ instead of $x$ in the first term and the input value can also be given as $3x$ instead of $x$ in the second term of the mathematical expression.
$=\,\,$ $\displaystyle 5 \times \large \lim_{5x\,\to\,0}{\normalsize \dfrac{\sin{5x}}{5x}}$ $-$ $\displaystyle 3 \times \large \lim_{3x\,\to\,0}{\normalsize \dfrac{\sin{3x}}{3x}}$
The second factor in each term is exactly same as the sine function based trigonometric limit rule.
Therefore, the limit of sine angle $5$ times $x$ divided $5$ times $x$ as the value of $5$ times $x$ approaches zero is equal to one. Similarly, the limit of sine angle $3$ times $x$ divided $3$ times $x$ as the value of $3$ times $x$ tends to zero is also equal to one.
$=\,\,$ $5 \times 1$ $-$ $3 \times 1$
Now, simplify the arithmetic expression to find the limit of the given trigonometric rational function as the value of $x$ is closer to $0$.
$=\,\,$ $5-3$
$=\,\,$ $2$
The limit can also be evaluated in the following two methods.
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