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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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I have added a homework tag per faq. Please edit to remove if this is not a homework. Anyhow, what have you tried about solving this problem ? –  Sasha Feb 6 '12 at 16:45
    
Do you know that $\lim_{t \to 0} \frac{\sin(t)}{t} = 1$? –  JavaMan Feb 6 '12 at 16:46
    
Yes, it is part of homework. First, I'll explain that its been ages since I have been in a math course. So I am very rusty on my algebra skills, and I am relearning identities. Now, to answer your question I am stuck at Sinpix /1 ( Cos sqrt3x / sin sqrt 3x). And here I am. –  tackyshirt Feb 6 '12 at 16:51
    
If it is necessary, please change the values for explanation purposes. I am so very rusty. –  tackyshirt Feb 6 '12 at 16:52
    
Looks like a poster child for L'Hôpital's rule. –  Henning Makholm Feb 6 '12 at 18:47

3 Answers 3

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.

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Thank you so much."One way to do that is to multiply by "1" in a helpful way" is what I really needed. I was referencing my notes and it skipped that step. –  tackyshirt Feb 6 '12 at 17:09
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In the last step, you could have written simply $\displaystyle\frac{\sin(\pi x)}{\pi x} \cdot \frac{\sqrt{3} x}{\sin(\sqrt{3} x)} \cdot \frac{\pi}{\sqrt{3}} \cos(\sqrt{3} x)$. –  Michael Hardy Feb 6 '12 at 17:41
    
@MichaelHardy Of course, but I would imagine the OP can handle it. –  JavaMan Feb 6 '12 at 17:43

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 $\sin'(0)=\tan'(0)=1$. –  Henning Makholm Feb 6 '12 at 19:04

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