# How do you find the Limit of [(sin x) * (sin^2 x)] / [1 -( cos x)] as x approaches 0?

Jun 20, 2016

Perform some conjugate multiplication and simplify to get ${\lim}_{x \to 0} \frac{\sin x \cdot {\sin}^{2} x}{1 - \cos x} = 0$

#### Explanation:

Direct substitution produces indeterminate form $\frac{0}{0}$, so we'll have to try something else.

Try multiplying $\frac{\sin x \cdot {\sin}^{2} x}{1 - \cos x}$ by $\frac{1 + \cos x}{1 + \cos x}$:
$\frac{\sin x \cdot {\sin}^{2} x}{1 - \cos x} \cdot \frac{1 + \cos x}{1 + \cos x}$

$= \frac{\sin x \cdot {\sin}^{2} x \left(1 + \cos x\right)}{\left(1 - \cos x\right) \left(1 + \cos x\right)}$

$= \frac{\sin x \cdot {\sin}^{2} x \left(1 + \cos x\right)}{1 - {\cos}^{2} x}$

This technique is known as conjugate multiplication, and it works nearly every time. The idea is to use the difference of squares property $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$ to simplify either the numerator or denominator (in this case the denominator).

Recall that ${\sin}^{2} x + {\cos}^{2} x = 1$, or ${\sin}^{2} x = 1 - {\cos}^{2} x$. We can therefore replace the denominator, which is $1 - {\cos}^{2} x$, with ${\sin}^{2} x$:
$\frac{\left(\sin x\right) \left({\sin}^{2} x\right) \left(1 + \cos x\right)}{{\sin}^{2} x}$

Now the ${\sin}^{2} x$ cancels:
$\frac{\left(\sin x\right) \left(\cancel{{\sin}^{2} x}\right) \left(1 + \cos x\right)}{\cancel{{\sin}^{2} x}}$
$= \left(\sin x\right) \left(1 + \cos x\right)$

Finish by taking the limit of this expression:
${\lim}_{x \to 0} \left(\sin x\right) \left(1 + \cos x\right)$
$= {\lim}_{x \to 0} \left(\sin x\right) {\lim}_{x \to 0} \left(1 + \cos x\right)$
$= \left(0\right) \left(2\right)$
$= 0$