# Homework Help: Question on solving complex limit with taylor

1. Jan 10, 2008

### transgalactic

question on solving a very hard limit with taylor

i want to solve this limit:

(1 - (cos x)^(sin(x^2)) ) / (sin (x^4))

i tried to solve it by tailor
the problem is that when i subtitute each trigonometric function with
a taylor series

i get a series in a power of a series
i dont know how to handle that

i generaly i dont know till what lenght to make each series
??

Last edited: Jan 10, 2008
2. Jan 11, 2008

### Gib Z

Ok first of all; What is your limit of? as (sin^2 x) approaches 15? Or as x approaches infinity?

Second of all, not all limits can be done with Taylor series...there are other methods that can do this limit though.

3. Jan 11, 2008

### transgalactic

for what cases we solve with tailor then

4. Jan 11, 2008

### transgalactic

anyone??

5. Jan 12, 2008

### Gib Z

Sigh you just have to learn from experience and realise when you can and when you can't. To get used to it, just try to use taylor series for everything for a while. If you try it for this one, it obviously dosent help.

6. Jan 12, 2008

### unplebeian

I am going to guess that it is approaching zero. Use L'Hopitals rule. Unless all other methods fail, try Taylor series at the end. Avoid Taylor series when the powers are trig functions. They might become messy.

7. Jan 12, 2008

### Gib Z

unplebeian: are you guessing that the OP meant that the value of the limit is approaching zero, or that x is approaching zero?

The OP still hasnt stated it >.> It the limit does not exist either way though. It oscillates horrendously around 0 and periodically afterwards. The OP choose a really bad example to begin limit how to evaluate limits on :(

Go to the Wikipedia articles http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule and http://en.wikipedia.org/wiki/Taylor_series to get more familiar with these things.

Last edited: Jan 12, 2008
8. Jan 12, 2008

### unplebeian

Yeah, I said in my post that I was guessing.

9. Jan 12, 2008

### Gib Z

No No i was asking for which one you were guessing at. 1) That x is approaching 0, or 2) That the value of the limit is approaching 0.

10. Jan 12, 2008

### transgalactic

its x approaching zero

11. Jan 13, 2008

### Gib Z

That limit oscillates horribly around zero, the limit doesn't exist. If you want an exercise with computing limits with Taylor series, show: $$\lim_{x\to 0} \frac{\sin x}{x} = 1, \lim_{x\to 0} \frac{\cos h -1}{h} = 0, \lim_{x\to 0} \frac{e^x -1}{x} = 0$$.

For more practice with Taylor series on things other than limits, show $$e^{ix} = \cos x + i \sin x$$, $$\sin (x+h) = \sin(x) \cos(h)+\cos(x) \sin(h)$$ and $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} ..., |x| < 1$$.

12. Jan 13, 2008

### olgranpappy

mmm. I think not.

13. Jan 13, 2008

### nicksauce

According to maple the limit does exist. But as I am not a masochist, I will not try to find out how to get the answer.

14. Jan 13, 2008

### olgranpappy

according to MMA it's 1/2...

15. Jan 13, 2008

### nicksauce

Actually, I may have found a solution with taylor:

$$cos(x)^{sin{x^2}}$$
~=
$$(1 - x^2/2)^{x^2}$$
~=
$$1 - \frac{x^4}{2}$$

So

$$\frac{1 - cos(x)^{sin(x^2)}}{sin(x^4)}$$
~=
$$\frac{x^4}{2sin(x^4)}$$
~= 1/2

Granted there are probably tons of unjustified simplifications here. A physicist's proof, not a mathematician's.

Last edited: Jan 13, 2008
16. Jan 13, 2008

### olgranpappy

A physicist's proof is the best kind, in my opinion. 1 - cos(x)^(sin(x^2)) does indeed behave near zero as (x^4)/2, and sin(x^4) behaves near zero as x^4. Good job.

17. Jan 13, 2008

### Gib Z

Yes this is indeed odd. Finding roots of the derivative of the function, there are infinitely many as we approach zero ie the derivative changes sign infinitely often near zero iee the function alternatives between increasing and decreasing infinitely often near zero. This was done with Maple by the way. Then, I asked it to find the limit, Maple states that it doesn't exist. I concluded that it was because of the oscillations, since right hand and left hand limits must be the same, the function being an even one. I even graphed it to check, it certainly looked like the limit was 1/2, but when i zoomed in I saw the oscillations.

However, when I saw these recent posts, I bothered to work it out myself in a similar manner to nicksauce, though with the mathematicians version, and I also conclude the limit is a half. Sorry =]

PS To Make that proof perfectly fine, just use Big-Oh notation. To make it a proper Physicists proof, denote the approximations with equalities =] And good work on not stating the limit on the last line.