# How do you find the limit of sin(x^2−4)/(x−2)  as x approaches 2?

Dec 14, 2016

Write it in a form that allows us to use ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$

#### Explanation:

It looks like we want $\theta = {x}^{2} - 4$, so we'll multiply by $\frac{x + 2}{x + 2}$, to get

$\frac{\left(x + 2\right) \sin \left({x}^{2} - 4\right)}{{x}^{2} - 4}$

${\lim}_{x \rightarrow 2} \sin \frac{{x}^{2} - 4}{x - 2} = {\lim}_{x \rightarrow 2} \left(x + 2\right) \cdot {\lim}_{x \rightarrow 2} \sin \frac{{x}^{2} - 4}{{x}^{2} - 4}$

$= \left(2 + \left(2\right)\right) \cdot \left(1\right) = 4$

Here is the graph.

graph{sin(x^2-4)/(x-2) [-4.305, 8.185, -1.205, 5.04]}

Dec 14, 2016

${\lim}_{x \to 2} \sin \frac{{x}^{2} - 4}{x - 2} = 4$

#### Explanation:

Substitute $t = x - 2$

$\sin \frac{{x}^{2} - 4}{x - 2} = \sin \frac{\left(x + 2\right) \left(x - 2\right)}{x - 2} = \sin \frac{t \left(t + 4\right)}{t} = \sin \frac{{t}^{2} + 4 t}{t} = \frac{\sin {t}^{2} \cos 4 t + \cos {t}^{2} \sin 4 t}{t} = t \cos 4 t \sin {t}^{2} / {t}^{2} + 4 \cos {t}^{2} \frac{\sin 4 t}{4 t}$

${\lim}_{x \to 2} \sin \frac{{x}^{2} - 4}{x - 2} = {\lim}_{t \to 0} t \cos 4 t \sin {t}^{2} / {t}^{2} + 4 \cos {t}^{2} \frac{\sin 4 t}{4 t} = 0 \cdot 1 \cdot 1 + 4 \cdot 1 \cdot 1 = 4$