Limit problem: lim x-> 0 ((tan(2x)^3/(x^3))

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kaebun

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I'm realy baffled by this problem.

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

I was wondering if someone could solve it out for me, and explain it. In my AP Review book it says to use lim x-> 0 (sinx/x)= 1 to solve it, and gives an example on how to solve it that way...

lim x->0 (((sin(2x)^3)/(x^3))(1/(cos(2x)^3))

lim x->0 (((8 (sin(2x)^3))/(2x^3))(1/(cos(2x)^3)) = 8(1)(1)=8

this is where I get confused, why would you choose to multiply the top and botom by 8 and not 4 or 3 or any other number? All I can figure is that maybe it has something to do with 2^3=8...


Thanks for any help
 
Do you HAVE to use the sinx/x thing?. We can use L'Hopital's rule.

\(\displaystyle \L\\\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}\)

\(\displaystyle \L\\f(x)=tan(2x); \;\ f'(x)=2+2tan^{2}(2x)\)

\(\displaystyle \L\\g(x)=x^{3}; \;\ g'(x)=3x^{2}\)

So, we have:

\(\displaystyle \L\\\frac{2}{3}\lim_{x\to\0}\frac{1+tan^{2}(2x)}{x^{2}}\)

\(\displaystyle \L\\\frac{2}{3}\lim_{x\to\0}[1+\underbrace{tan^{2}(2x)}_{\text{limit is 0}}]\lim_{x\to\0}\frac{1}{x^{2}}\)

\(\displaystyle \L\\\frac{2}{3}(1)\lim_{x\to\0}\frac{1}{x^{2}}={\infty}\)

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

But you could rewrite, since \(\displaystyle \L\\tan(2x)=\frac{sin(2x)}{cos(2x)}\)

and

\(\displaystyle \L\\\sin(2x)=2sin(x)cos(x)\)

\(\displaystyle \L\\\frac{2cos(x)}{cos(2x)}\cdot\frac{sin(x)}{x^{3}}\)

\(\displaystyle \L\\2\lim_{x\to\0}\frac{cos(x)}{cos(2x)}=2\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{x^{3}}={\infty}\)
 
No, I actualy tried to solve it using l'Hopitals rule, and it got all icky, and weird.
Why does f'(x)= 2+2 tan(2x)^2 ? I thought the derivitive of tan(u)= sec(u)^2 du . Everything else you did made sense but infinity isn't one of the options, and the book says 8 is supposed to be the answer
 
The derivative of tan(2x) is \(\displaystyle \L\\\frac{2}{cos^{2}(2x)}=2(1+tan^{2}(2x))=2sec^{2}(2x)\).

I just used \(\displaystyle 2+2tan^{2}(2x)\)

I don't know where they get 8?. It's infinity.

For that matter, enter in some values close to 0 and you can see it gets larger and larger. There is no limit.
 
The easiest way to calculate limits is by using series expansions. In this case you could use that \(\displaystyle \L\tan(x) = x + \frac{x^{3}}{3} +O(x^5)\)

where \(\displaystyle \L O(x^5)\) is a function that is smaller than some constant times \(\displaystyle \L x^5\) in some interval containing the point x = 0. You see that if the limit involving the tan function is to have a finite limit then you must subtract the term linear in x. E.g. from the above series expansion, you see that

\(\displaystyle \L\frac{\tan(2x)-2x}{x^3}= \frac{8}{3} + O(x^2)\)

And the limit of the left hand side for x to zero is thus 8/3.
 
I see what's going on. You have a typo in your problem statement.

No wonder we're scratching our heads. You forgot the cube in the tan(2x).

It is supposed to be :

\(\displaystyle \L\\\lim_{x\to\0}\frac{tan^{3}(2x)}{x^{3}}\)

\(\displaystyle \L\\tan^{3}(2x)=\frac{sin^{3}(2x)}{cos^{3}(2x)}\)

But, \(\displaystyle \L\\sin^{3}(2x)=8sin^{3}(x)cos^{3}(x)\)

Rewrite as \(\displaystyle \L\\\lim_{x\to\0}\frac{8sin^{3}(x)cos^{3}(x)}{x^{3}cos^{3}(2x)}\)

\(\displaystyle \L\\8\lim_{x\to\0}\frac{sin^{3}(x)}{x^{3}}\cdot\lim_{x\to\0}\frac{cos^{3}(x)}{cos^{3}(2x)}\)

\(\displaystyle \L\\8(1)(1)=8\)
 
Oops! Sorry about that!
Thank you for your help, I think I understand it now
 
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