QUESTION

# Solve the following: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{x}$

Hint: Here we need to convert the given expression into any standard formulae of the limits such that the simplification is easier. Here we will use $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ to evaluate.

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{x}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{3\sin 3x}}{{3x}}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
$\Rightarrow 3 \times 1 = 3$