Trig limit: lim [tan(3x)] / [3 tan(2x)] as x goes to zero

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ku1005

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Oct 27, 2006
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I have been working on the following limit for ages, it seems, and I just don't seem to be getting anywhere -- which ususally means I am missing something obvious. Any pointers would be greatly apprecited!

lim x-> 0 for tan(3x) / 3tan(2x)

I first tried the obvious route, substitution, getting a "value" of 0/0, which is no help. Then I converted the tangents into sines over cosines, and tried to work with their respective x-values. But I was unable to remove the sine from the denominator, meaning that I would always have something divided by zero. Hints?

Also, do you know of any sites with pointers for solving trig identities, like techniques?

With any luck, I'll have this figured out by the time somebody replies. In either case, thank you for looking at this, and thank you for any help you can offer.
 
tan3x / 3 tan 2x lim x-->0 0/0 undefined by L'Hospitals rule take derivatives
1/3 [ tan3x / tan 2x]
{1/3} { 1/cos^23x / 1/cos^22x}
{1/3} { cos2x/cos3x}^2 lim x-->0
1/3 answer

please check for errors
Arthur
 
\(\displaystyle \L \lim_{x \rightarrow 0} \frac{\tan{(3x)}}{3\tan{(2x)}} =\)

\(\displaystyle \L \lim_{x \rightarrow 0} \frac{x\tan{(3x)}}{3x\tan{(2x)}} =\)

\(\displaystyle \L \lim_{x \rightarrow 0} \frac{\tan{(3x)}}{3x} \cdot \frac{x}{\tan{(2x)}} =\)

\(\displaystyle \L \lim_{x \rightarrow 0} \frac{\tan{(3x)}}{3x} \cdot \lim_{x \rightarrow 0}\frac{x}{\tan{(2x)}} =\)

\(\displaystyle \L \lim_{x \rightarrow 0} \frac{\tan{(3x)}}{3x} \cdot \frac{1}{2}\lim_{x \rightarrow 0}\frac{2x}{\tan{(2x)}} =\)

\(\displaystyle \L \lim_{x \rightarrow 0} \frac{\sin{(3x)}}{3x\cos{(3x)}} \cdot \frac{1}{2}\lim_{x \rightarrow 0}\frac{2x\cos{(2x)}}{\sin{(2x)}} =\)

\(\displaystyle \L \lim_{x \rightarrow 0} \frac{\sin{(3x)}}{3x}\frac{1}{\cos{(3x)}} \cdot \frac{1}{2}\lim_{x \rightarrow 0}\frac{2x}{\sin{(2x)}}\cos{(2x)} =\)

\(\displaystyle \L \lim_{3x \rightarrow 0} \frac{\sin{(3x)}}{3x}\frac{1}{\cos{(3x)}} \cdot \frac{1}{2}\lim_{2x \rightarrow 0}\frac{2x}{\sin{(2x)}}\cos{(2x)} =\)

\(\displaystyle \L 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{2}\)
 
Re: Trig Limit...im having troubly solving...plz help!

Hello, ku1005!

\(\displaystyle \L\lim_{x\to0}\,\frac{\tan(3x)}{3\tan(2x)}\)
Click to expand...

Since it goes to \(\displaystyle \frac{0}{0}\), we can use L'Hopital's Rule.

We have: \(\displaystyle \L\:\lim_{x\to0}\,\frac{3\sec^2(3x)}{3\cdot2\sec^2(2x)}\;\;\Rightarrow\;\;\frac{3\sec^2(0)}{6\sec^2(0)} \;=\;\frac{3}{6}\;=\;\frac{1}{2}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This can be done "the long way" with this theorem: . \(\displaystyle \L\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1\)


We have: \(\displaystyle \L\:\frac{1}{3}\cdot\frac{\tan(3x)}{\tan(2x)} \;=\;\frac{1}{3}\cdot\frac{\sin(3x)}{\cos(3x)}\cdot\frac{\cos(2x)}{\sin(2x)} \;=\;\frac{1}{3}\cdot\frac{\cos(2x)}{\cos(2x)}\cdot\frac{\sin(3x)}{1}\cdot\frac{1}{\sin(2x)}\)

. . \(\displaystyle \L= \;\frac{1}{3}\cdot\frac{\cos(2x)}{\cos(3x)}\cdot\left(\frac{3x}{3x}\cdot\frac{\sin(3x)}{1}\right)\left(\frac{2x}{2x}\cdot\frac{1}{\sin(2x)\right) \;=\;\frac{1}{3}\cdot\frac{\cos(2x)}{\cos(3x)}\cdot\frac{3x}{2x}\cdot\frac{\sin(3x)}{3x}\cdot\frac{2x}{\sin(2x)\)

And we have: \(\displaystyle \L\;\,\frac{1}{2}\cdot\frac{\cos(2x)}{\cos(3x)}\cdot\frac{\sin(3x)}{3x}\cdot\frac{2x}{\sin(2x)}\)
. . . . . . . . . . . .\(\displaystyle \downarrow\;\;\,\downarrow\;\;\;\:\:\:\downarrow\;\;\;\;\,\downarrow\)
Take the limit: \(\displaystyle \L\;\frac{1}{2}\;\cdot\;\frac{1}{1}\qquad\:\cdot\qquad\: 1 \quad\:\cdot\quad\: 1 \;\;=\;\;\fbox{\frac{1}{2}}\)

Too fast for me, skeeter!
 
thanks heaps for all your help guys...really appreciate it...just dunno how u do it so fast lol

but thanks again
 
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