# Question 368ef

Oct 18, 2017

The limit is $0$

#### Explanation:

lim_(x->0) (sinx(1-cosx))/(xtan(pix)

Separate this into a product like this:

$= {\lim}_{x \to 0} \left(\sin \frac{x}{x}\right) \left(\frac{1 - \cos x}{\tan} \left(\pi x\right)\right)$

$= {\lim}_{x \to 0} \left(\sin \frac{x}{x}\right) \cdot {\lim}_{x \to 0} \frac{1 - \cos x}{\tan} \left(\pi x\right)$

Use the limit of sin(x) / x rule:

$= 1 \cdot {\lim}_{x \to 0} \frac{1 - \cos x}{\tan} \left(\pi x\right)$

Now use the trig rule $1 - \cos x = 2 {\sin}^{2} \left(\frac{x}{2}\right)$

$= {\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(\frac{x}{2}\right)}{\tan} \left(\pi x\right)$

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Since this limit approaches 0, we can see that both $\frac{x}{2}$ and $\pi x$ approach $0$ as well. This means that at extremely small values of $x$, we can essentially say that $x \approx \frac{x}{2}$ and $x \approx \pi x$, so we can say that:

${\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(\frac{x}{2}\right)}{\tan} \left(\pi x\right) = {\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(x\right)}{\tan} \left(x\right)$

$= {\lim}_{x \to 0} 2 \sin x \cos x$

$= 2 \sin \left(0\right) \cos \left(0\right)$

$= 2 \left(0\right) \left(1\right) = 0$

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So the limit equals $0$. This can be confirmed with a graphing calculator:
graph{(sinx(1-cosx))/(xtan(pix)) [-1.25, 1.25, -0.625, 0.625]}

Oct 18, 2017

#### Explanation:

$\frac{\sin \left(1 - \cos x\right)}{x \tan \left(\pi x\right)} = \sin \frac{1 - \cos x}{\left(1 - \cos x\right)} \cdot \frac{1 - \cos x}{x \tan \left(\pi x\right)}$

Since, ${\lim}_{x \rightarrow 0} \sin \frac{1 - \cos x}{1 - \cos x} - {\lim}_{u \rightarrow 0} \sin \frac{u}{u} = 1$, we will concentrate on the second factor.

((1-cosx))/(xtan(pix)) * ((1+cosx))/((1+cosx)) = sin^2x/(xtan(pix) ((1+cosx))#

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

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

Taking the limit as $x \rightarrow 0$ gives

$= 1 \cdot \frac{1}{\pi} \cdot 1 \cdot \frac{1}{1 + 1} = \frac{1}{2 \pi}$