Finding a limit using power series expansion

[Daniel Lobo]

Homework Statement

The problem wants me to find the limit below using series expansion.
$\lim_{x \to 0}(\frac{1}{x^2}\cdot \frac{\cos x}{(\sin x)^2})$

Homework Equations

The Attempt at a Solution

(1) For startes I'll group the two fractions inside the limit together
$\lim_{x \to0}\frac{sin^2(x)-x^2cos(x)}{x^2sin^2(x)}$[/B]
This limit is of the form 0/0 so I can use l'hopital's rule here, I'll expand both the numerator and the denominator into their respective power series and take enough derivatives so that the functions don't equal to zero at x=0.
(2) The numerator expanded:
$sin^2(x)-x^2cos(x)=\sum_{n=0}^\infty\frac{x^{4n+2}}{((2n+1)!)^2}-\sum_{n=0}^\infty\frac{(-1)^{n}x^{2n+2}}{(2n)!}=(x^2+\frac{x^6}{36}+...)-(x^2-\frac{x^4}{2}+...)=(\frac{x^4}{2}+\frac{x^6}{36}+...)$
(3) The x^(4)/2 is the only part of interest to me since it's the only part that will not be zero after taking 4 derivatives and setting x=0 and doing so will wield:
$\frac{\mathrm{d^4} }{\mathrm{d} x^4}(sin^2(x)-x^2cos(x))|_{x=0}=12$


The denominator expanded:
$x^2sin^2(x)=\sum_{n=0}^\infty\frac{x^{4n+4}}{(2n+1)^2!}=(x^4+\frac{x^8}{4}+...)$
(4) Doing the same as in (3):
$\frac{\mathrm{d^4} }{\mathrm{d} x^4}(x^2sin^2(x))|_{x=0}=24$
With these results I reach my answer of 12/24=1/2 but my book says the answer is 1/6 and I can't figure out what I'm doing wrong for the life of me, it's been 3 hours on this exercise already and would really apreciate some advice. Also apologies for the iffy english.

 

[pasmith]

Homework Statement

The problem wants me to find the limit below using series expansion.
$\lim_{x \to 0}(\frac{1}{x^2}\cdot \frac{\cos x}{(\sin x)^2})$

Homework Equations

The Attempt at a Solution

(1) For startes I'll group the two fractions inside the limit together
$\lim_{x \to0}\frac{sin^2(x)-x^2cos(x)}{x^2sin^2(x)}$

The operator in the problem statement is the \cdot operator; that means multiplication, not subtraction.

This limit is of the form 0/0 so I can use l'hopital's rule here, I'll expand both the numerator and the denominator into their respective power series and take enough derivatives so that the functions don't equal to zero at x=0.

(2) The numerator expanded:
$sin^2(x)-x^2cos(x)=\sum_{n=0}^\infty\frac{x^{4n+2}}{((2n+1)!)^2}-\sum_{n=0}^\infty\frac{(-1)^{n}x^{2n+2}}{(2n)!}=(x^2+\frac{x^6}{36}+...)-(x^2-\frac{x^4}{2}+...)=(\frac{x^4}{2}+\frac{x^6}{36}+...)$
(3) The x^(4)/2 is the only part of interest to me since it's the only part that will not be zero after taking 4 derivatives and setting x=0 and doing so will wield:
$\frac{\mathrm{d^4} }{\mathrm{d} x^4}(sin^2(x)-x^2cos(x))|_{x=0}=12$The denominator expanded:
$x^2sin^2(x)=\sum_{n=0}^\infty\frac{x^{4n+4}}{(2n+1)^2!}=(x^4+\frac{x^8}{4}+...)$
(4) Doing the same as in (3):
$\frac{\mathrm{d^4} }{\mathrm{d} x^4}(x^2sin^2(x))|_{x=0}=24$
With these results I reach my answer of 12/24=1/2 but my book says the answer is 1/6 and I can't figure out what I'm doing wrong for the life of me, it's been 3 hours on this exercise already and would really apreciate some advice. Also apologies for the iffy english.

Your task is greatly eased by means of the identity <br /> \sin^2 x = \tfrac12(1 - \cos 2x) which eliminates the need to square the series for sine, which you have done incorrectly. The square of \sum_{n=0}^\infty a_nx^n is <br /> \sum_{n=0}^\infty \sum_{m=0}^\infty a_na_mx^{n+m} = \sum_{n=0}^\infty \left(\sum_{k = 0}^n a_k a_{n-k}\right) x^n and not <br /> \sum_{n=0}^\infty a_n^2 x^{2n} as you have.

 

[Daniel Lobo]

Thank you I understand where I made my mistake in assuming the sin²x expansion, and \cdot operator is seems to have gone unnoticed. Thanks again.

 

[SammyS]

Homework Statement

The problem wants me to find the limit below using series expansion.
$\lim_{x \to 0}(\frac{1}{x^2}\cdot \frac{\cos x}{(\sin x)^2})$

Homework Equations

The Attempt at a Solution

(1) For startes I'll group the two fractions inside the limit together
$\displaystyle \lim_{x \to0}\frac{sin^2(x)-x^2cos(x)}{x^2sin^2(x)}$[/B]
...

There appears to be an error in your statement of the problem. If the problem is as you state it, then the limit is ∞. Not that difficult to show.

Judging by that next equation it appears that the problem statement should be:

Find the limit using series expansion.
$\displaystyle \lim_{x \to 0}\left(\frac{1}{x^2}-\frac{\cos x}{\sin^2x}\right)$​

Use the expression for sin²(x) given by pasmith, then use the expansion for cosine:

$\displaystyle \cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}-\dots$​

.