# How do you find the limit of (sin12x)/x as x approaches zero?

Aug 9, 2015

Use L'Hôpital's rule with $f \left(x\right) = \sin \left(12 x\right)$ and $g \left(x\right) = x$.

$f ' \left(x\right) = 12 \cos 12 x$ and $g ' \left(x\right) = 1$, so

${\lim}_{x \to 0} f \frac{x}{g} \left(x\right) = {\lim}_{x \to 0} \frac{f ' \left(x\right)}{g ' \left(x\right)} = {\lim}_{x \to 0} \frac{12 \cos 12 x}{1} = 12$

#### Explanation:

L'Hôpital's rule says that if:

(1) $f \left(x\right)$ and $g \left(x\right)$ are differentiable on an open interval $I$ containing $c$, except possibly at $c$ and

(2) ${\lim}_{x \to c} f \left(x\right) = {\lim}_{x \to c} g \left(x\right) = 0$ and

(3) ${\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ exists and $g ' \left(x\right) \ne 0$ for all $x \in I$ (except possibly at $c$)

then

${\lim}_{x \to c} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

With $f \left(x\right) = \sin \left(12 x\right)$ and $g \left(x\right) = x$ these 3 conditions hold for $c = 0$.