According to the limit by direct substitution method, the limit of sine of angle $5$ times $x$ minus sine of angle $3$ times $x$ divided by $x$ as the value of variable $x$ tends to zero is indeterminate.
$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}-\sin{3x}}{x}}$ $\,=\,$ $\dfrac{0}{0}$
The indeterminate form interrupts us to proceed further. The difference between the sine functions in numerator, and the variable function in denominator are the main reason for this indeterminate form.
The L’Hospital’s rule is really useful to find the limit of this rational function by eliminating the indeterminate form from the rational function with differentiation.
So, let’s use the l’hospital’s rule to eliminate the indeterminate form from the rational function.
$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{5x}-\sin{3x})’}{(x)’}}$
The expressions in both numerator and denominators are defined in terms of $x$. So, they both should be differentiated with respect to $x$ to use the L’Hôpital’s rule.
$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}(\sin{5x}-\sin{3x})}{\dfrac{d}{dx}\,(x)}}$
It is time to find the differentiation of the expression in the numerator and also the derivative of the expression in the denominator.
Look at the trigonometric expression in the numerator. According the difference rule of the derivatives, the derivative of the difference between the trigonometric functions can be calculated by the difference of their derivates.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}\,\sin{5x}-\dfrac{d}{dx}\,\sin{3x}}{\dfrac{dx}{dx}}}$
According to the derivative rule of sine of a multiple angle, The derivative of sine of angle five times $x$ with respect to $x$ is equal to $5$ times cosine of angle five times $x$, and the derivative of sine of angle three times $x$ with respect to $x$ is equal to three times cosine of angle $5$ times $x$.
The derivative of $x$ with respect to $x$ is equal one as per the derivative rule of a variable.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{5 \times \cos{5x}-3 \times \cos{3x}}{1}}$
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize (5 \times \cos{5x}-3 \times \cos{3x})}$
The differentiation in the L’Hospital’s rule transformed the rational function into a non-rational function. Now, it is possible to find the limit without indeterminate form.
So, let’s find the limit by direct substitution method as the value of $x$ is closer to zero.
$=\,\,$ $5 \times \cos{5(0)}$ $-$ $3 \times \cos{3(0)}$
$=\,\,$ $5 \times \cos{(5 \times 0)}$ $-$ $3 \times \cos{(3 \times 0)}$
$=\,\,$ $5 \times \cos{(0)}$ $-$ $3 \times \cos{(0)}$
$=\,\,$ $5 \times \cos{0}$ $-$ $3 \times \cos{0}$
The cosine of zero radian is equal to one as per the trigonometry and substitute it in the above trigonometric expression.
$=\,\,$ $5 \times 1-3 \times 1$
Finally, simplify the arithmetic expression to find the limit.
$=\,\,$ $5-3$
$=\,\,$ $2$
The limit problem can also be solved in the following two methods.
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