Trig limit question involving cot

How would I solve the following trig limit

$$\lim_{x\rightarrow 0} \frac{4x}{\cot 3x}$$

This is what I did

$$\frac{4}{3} \frac{\sin3x}{\cos3x}$$

$$\frac{4}{3} (\frac{\sin3x}{3x}) (\frac{\cos3x}{1})$$

limit equals $\frac{4}{3}$ would this be correct.

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it is an hw question but I am interested in hearing the detail... –  Fernando Martinez Jan 29 '13 at 17:06
@FernandoMartinez: I formatted your questions using MathJax. Please make sure I got it correct. Regards –  Amzoti Jan 29 '13 at 17:07
@Amazoti it is correct. Indeed –  Fernando Martinez Jan 29 '13 at 17:08
Your answer is incorrect. Regards –  Amzoti Jan 29 '13 at 17:09

$$\lim_{x\to 0} \frac{4x}{\cot 3x} = \lim_{x\rightarrow 0}\, (4x) \frac{\sin 3x}{\cos 3x} =4\cdot 0 \cdot \frac 01 = 0.$$

The main problem you had was the result of a few mistakes in the your algebra when manipulating the function.

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Fernando - is this clear now? Your main problem was a few mistakes in you algebra, which led to your evaluation of a different limit than the one given. –  amWhy Jan 29 '13 at 18:17
Hmm I know I did it incorrect now but I am not sure where I made my error. –  Fernando Martinez Jan 30 '13 at 18:56
your $\dfrac{\sin 3x}{\cos 3x}$ term was correct, but you forgot the $x$ from the original equation, and you introduced the fraction $\dfrac 43$ which should have been just the original $4$. So we need $(4x)\dfrac{\sin 3x}{\cos 3x}$. –  amWhy Jan 30 '13 at 19:01
$\dfrac{4x}{\cot 3x} = 4x \tan 3x = (4x) \dfrac{\sin 3x}{\cos 3x}$ or alternatively, $\dfrac{4x}{\cot 3x} = \dfrac{4x}{\dfrac{\cos 3x}{\sin 3x}} = \dfrac{4x \sin 3x}{\cos 3x} = 4x\cdot \dfrac{\sin 3x}{\cos 3x}$ –  amWhy Jan 30 '13 at 19:03
Fernando, is this any clearer now? All that was done, before evaluating the limit, was rewriting the function to an equivalent form which allowed us to find the limit. –  amWhy Jan 30 '13 at 19:11

$$\lim_{x\rightarrow 0} \frac{4x}{\cot 3x} = \lim_{x\rightarrow 0} 4 x \tan{3x} = \lim_{x\rightarrow 0} (4 x) (3 x) = 0$$.

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Why make this step $\lim_{x\rightarrow 0} 4 x \tan{3x} = \lim_{x\rightarrow 0} (4 x) (3 x)$? The LHS is just $0\cdot0$. –  Git Gud Jan 29 '13 at 17:16
To help show where the OP made a mistake. He had a fraction flipped and therefore got a constant. –  Ron Gordon Jan 29 '13 at 17:19
oohh! Well done. –  Git Gud Jan 29 '13 at 17:20