Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}-\sin{3x}}{x}}$
The limit of sine of angle five times $x$ minus sine of angle $3$ times $x$ divided by $x$ as the value of $x$ approaches to zero should be evaluated in this limit problem.

So, let’s try to find its limit by the direction substitution method firstly.
Direct substitution
$=\,\,$ $\dfrac{\sin{5(0)}-\sin{3(0)}}{0}$
$=\,\,$ $\dfrac{\sin{(5 \times 0)}-\sin{(3 \times 0)}}{0}$
$=\,\,$ $\dfrac{\sin{(0)}-\sin{(0)}}{0}$
$=\,\,$ $\dfrac{\sin{0}-\sin{0}}{0}$
According to the trigonometry, the sine of zero radian is zero. Now, substitute it in each term of the numerator of the above rational trigonometric expression.
$=\,\,$ $\dfrac{0-0}{0}$
$=\,\,$ $\dfrac{0}{0}$
The zero divided by zero is an indeterminate form mathematically.
It clears that the limit by direct substitution method is failed to find the limit of sine of angle $5x$ minus sine of angle $3x$ divided by $x$ as the value of $x$ tends to zero.
Methods
The limit can be evaluated in three different methods possibly and let’s learn each method to find the limit of $\sin{5x}$ minus $\sin{3x}$ divided by $x$ as the value of $x$ is closer to zero.
- $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}-\sin{3x}}{x}}$Limit RulesLearn how to find the limit by using the Limits rules.
- $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}-\sin{3x}}{x}}$Trigonometric identitiesLearn how to find the limit by trigonometric identities.
- $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{5x}-\sin{3x}}{x}}$L’Hôpital’s RuleLearn how to find the limit by using the L’Hospital’s Rule.
