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$ \lim_{x \to 0} \frac{\sin(\pi x)}{\tan(\sqrt{3} x)} $

I need a step by step explanation. Thank you.

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    Looks like a $p$oster child for L'Hôpital's rule.2012-02-06

3 Answers 3

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Hint:

First, let's expand what we are looking at:

$\begin{align} \frac{\sin(\pi x)}{\tan(\sqrt{3}x)} = \frac{\sin(\pi x)}{\frac{\sin(\sqrt{3} x)}{\cos(\sqrt{3} x)}} = \sin(\pi x) \cdot \frac{\cos(\sqrt{3}x)}{\sin(\sqrt{3}x)} = \sin(\pi x)\cdot\frac{1}{\sin(\sqrt{3}x)}\cdot \cos(\sqrt{3}x). \end{align}$ Next, we want to get each $\sin$ term in the form $\frac{\sin(t)}{t}$. One way to do that is to multiply by "$1$" in a helpful way: $\begin{align} \sin(\pi x)\cdot\frac{1}{\sin(\sqrt{3}x)}\cdot \cos(\sqrt{3}x) &= \sin(\pi x)\frac{\pi x}{\pi x} \cdot \frac{1}{\sin(\sqrt{3}x)} \cdot \frac{\sqrt{3}x}{\sqrt{3} x} \cos(\sqrt{3}x) \\ &= \frac{\sin(\pi x)}{\pi x} \cdot \frac{\sqrt{3} x}{\sin(\sqrt{3} x)} \cdot \frac{\pi x}{\sqrt{3} x} \cos(\sqrt{3} x). \end{align}$

I'll let you take a limit and see what happens.

  • 0
    @MichaelHardy Of course, but I would imagine the OP can handle it.2012-02-06
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Results going to be used:

  • $\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} =1 = \lim_{x \to 0} \frac{\tan{x}}{x}$

What you have is \begin{align*} \lim_{x \to 0}\: \frac{\sin(\pi{x})}{\tan\sqrt{3}{x}} &= \lim_{x \to 0} \: \frac{\sin\pi{x}}{\pi{x}} \times \frac{\sqrt{3}x}{\tan{\sqrt{3}x}} \times \frac{\pi{x}}{\sqrt{3}x} \\ &= \lim_{x \to 0} \: \frac{\pi{x}}{\sqrt{3}x} = \frac{\pi}{\sqrt{3}} \end{align*}

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As you have a $0/0$ form, L'Hôpital's rule works nicely here: \eqalign{ \lim_{x\rightarrow 0}{\sin(\pi x)\over \tan(\sqrt3 x) } &=\lim_{x\rightarrow 0}{\bigl[ \sin(\pi x)\bigr]'\over \bigl[\tan(\sqrt3 x)\bigr]' }\cr &=\lim_{x\rightarrow 0}{ \cos(\pi x) \cdot [\pi x]'\over \sec^2(\sqrt3 x)\cdot\bigl[ \sqrt 3 x\bigr]' }\cr &=\lim_{x\rightarrow 0}{\pi\cos(\pi x)\over \sqrt 3\,\sec^2(\sqrt3 x) }\cr &=\lim_{x\rightarrow 0}\Bigl[{\pi \over \sqrt 3}\cos(\pi x)\,\cos^2(\sqrt3 x) \Bigr]\cr &= {\pi\over\sqrt3}.}

Of course, the other answers are nicer, as they are more elementary.

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    The other answers may be more elementary, but this is much, much faster, especially if one knows already that $\sin'(0)=\tan'(0)=1$.2012-02-06