# How do you find the limit of (3 sin (x)(1 - cos(x))) / (x^2)  as x approaches 0?

Oct 24, 2016

$L {t}_{x \to 0} \frac{3 \sin \left(x\right) \left(1 - \cos \left(x\right)\right)}{x} ^ 2 = 0$

#### Explanation:

To find $L {t}_{x \to 0} \frac{3 \sin \left(x\right) \left(1 - \cos \left(x\right)\right)}{x} ^ 2$, let us divide it in two parts

$L {t}_{x \to 0} \frac{3 \sin \left(x\right)}{x} \times L {t}_{x \to 0} \frac{1 - \cos \left(x\right)}{x}$

Now as each of these is of the form $\frac{0}{0}$ when $x \to 0$,

we can use L'Hospital's Rule,

and as such $L {t}_{x \to 0} \frac{3 \sin \left(x\right)}{x} \times L {t}_{x \to 0} \frac{1 - \cos \left(x\right)}{x}$

= $L {t}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} 3 \sin \left(x\right)}{1} \times L {t}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(1 - \cos \left(x\right)\right)}{1}$

= $L {t}_{x \to 0} \frac{3 \cos \left(x\right)}{1} \times L {t}_{x \to 0} \frac{0 - \left(- \sin \left(x\right)\right)}{1}$

= $3 \times \frac{0}{1} = 0$
graph{(3sin(x)(1-cos(x)))/x^2 [-2.5, 2.5, -1.25, 1.25]}

Oct 24, 2016

$0$

#### Explanation:

$1 - \cos \left(x\right) = \cos \left(0\right) - \cos \left(x\right)$

we know that

$\cos \left(a + b\right) - \cos \left(a - b\right) = - 2 \sin \left(a\right) \sin \left(b\right)$

calling

$\left\{\begin{matrix}a + b = 0 \\ a - b = x\end{matrix}\right.$

and solving for $a , b$

we have $a = \frac{x}{2} , b = - \frac{x}{2}$ so

$\cos \left(0\right) - \cos \left(x\right) = 2 {\sin}^{2} \left(\frac{x}{2}\right)$

then

$\frac{3 \sin \left(x\right) \left(1 - \cos \left(x\right)\right)}{x} ^ 2 = \frac{6 \sin \left(x\right) {\sin}^{2} \left(\frac{x}{2}\right)}{x} ^ 2$

Now using the fundamental result

${\lim}_{x \to 0} \sin \frac{x}{x} = 1$ we have

${\lim}_{x \to 0} \frac{3 \sin \left(x\right) \left(1 - \cos \left(x\right)\right)}{x} ^ 2 = \frac{6}{4} {\lim}_{x \to 0} \sin \left(x\right) {\lim}_{x \to 0} {\left(\sin \frac{\frac{x}{2}}{\frac{x}{2}}\right)}^{2} = \frac{3}{2} \cdot 0 \cdot 1 = 0$

Oct 24, 2016

We can use the two fundamental trigonometric limits.

#### Explanation:

We have

${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$ $\text{ }$ and $\text{ }$ ${\lim}_{x \rightarrow 0} \frac{1 - \cos x}{x} = 0$.

These limits, together with the product property of limits allow us to write:

${\lim}_{x \rightarrow 0} \frac{3 \sin x \left(1 - \cos x\right)}{x} ^ 2 = 3 {\lim}_{x \rightarrow 0} \left(\sin \frac{x}{x}\right) {\lim}_{x \rightarrow 0} \frac{1 - \cos x}{x}$

$= 3 \left(1\right) \left(0\right) = 0$