Find the limit using taylor series

[doktorwho]

Homework Statement

Using the taylor series at point $(x=0)$ also known as the meclaurin series find the limit of the expression:
$$L=\lim_{x \rightarrow 0} \frac{1}{x}\left(\frac{1}{x}-\frac{cosx}{sinx}\right)$$

Homework Equations

3. The Attempt at a Solution [/B]
$L=\lim_{x \rightarrow 0} \frac{1}{x}\left(\frac{sinx-xcosx}{xsinx}\right)$
$L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{sinx-xcosx}{sinx}\right)$
I need help to continue. This is the first problem I am doing that asks me to claculate the limit using the taylor series. Am i suppose to write the taylor series for $sinx,cosx$ up to some element and somehow use them?

 

[Orodruin]

Am i suppose to write the taylor series for $sinx,cosx$ up to some element and somehow use them?

Yes. Keep the leading non-zero term in $x$ in both denominator and numerator. The rest will be irrelevant.

 

[fresh_42]

Shorter is directly the series for cotangent.

 

[vela]

You may also find the series $\frac{1}{1-z} = 1+z+z^2+z^3+\cdots$ useful.

 

[vela]

Shorter is directly the series for cotangent.

Yeah, but who remembers that?

 

[fresh_42]

Yeah, but who remembers that?

Wiki does. The advantage is, you don't even need a pencil.

 

[doktorwho]

So taylor for $\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...$ and for $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}+...$
So:
$L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...)-x(1 - \frac{x^2}{2!} + \frac{x^4}{4!}+...)}{x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...}\right)$
So beside the x getting canceled in the upper part what else can i do?

 

[fresh_42]

So taylor for $\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...$ and for $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}+...$
So:
$L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...)-x(1 - \frac{x^2}{2!} + \frac{x^4}{4!}+...)}{x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ...}\right)$
So beside the x getting canceled in the upper part what else can i do?

Do what @Orodruin has said: calculate the difference, multiply with $\frac{1}{x^2}$, cancel the quotient such that leading term of the denominator equals $1$ and see which terms without an $x$ are left. (I haven't done it this way, though.)

 

[doktorwho]

Do what @Orodruin has said: calculate the difference, multiply with $\frac{1}{x^2}$, cancel the quotient such that leading term of the denominator equals $1$ and see which terms without an $x$ are left. (I haven't done it this way, though.)

Since the terms after $\frac{x^3}{3!}$ are getting smaller and smaller can i ignore them and write just the two leading members of the series?
Like:
$L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{(x - \dfrac{x^3}{3!}) -x(1 - \frac{x^2}{2!})}{x - \dfrac{x^3}{3!}}\right)$
$L=\lim_{x \rightarrow 0}\left(\frac{(x - \dfrac{x^3}{3!}) -x(1 - \frac{x^2}{2!})}{x^3}\right)$
$L=\lim_{x \rightarrow 0}\left(\frac{x^3}{3x^3}\right)=\frac{1}{3}$
Would this be correct?

 

[fresh_42]

Yes, it is correct. Mathematically one would have notated the missing terms like this: $\; \sin(x)=x-\frac{x^3}{3!}+O(x^5)$ instead of just dropping them, but the result is the same.

($O(x^5)$ means basically: "something with $x^5$ as factor".)

 

[Ray Vickson]

Homework Statement

Using the taylor series at point $(x=0)$ also known as the meclaurin series find the limit of the expression:
$$L=\lim_{x \rightarrow 0} \frac{1}{x}\left(\frac{1}{x}-\frac{cosx}{sinx}\right)$$

Homework Equations

3. The Attempt at a Solution [/B]
$L=\lim_{x \rightarrow 0} \frac{1}{x}\left(\frac{sinx-xcosx}{xsinx}\right)$
$L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{sinx-xcosx}{sinx}\right)$
I need help to continue. This is the first problem I am doing that asks me to claculate the limit using the taylor series. Am i suppose to write the taylor series for $sinx,cosx$ up to some element and somehow use them?

Yes. Keep the leading non-zero term in $x$ in both denominator and numerator. The rest will be irrelevant.

This is not quite right, and can sometimes lead to an incorrect answer (as it does in this example). If we keep only $1-x^2/2$ from $\cos x$ and just retain $x$ from $\sin x$ we get
$$\text{function} \doteq \frac{1}{x} \left( \frac{1}{x} - \frac{1}{x} + \frac{x^2}{2x} \right) = \frac{1}{2}.$$
This would lead to the incorrect limit 1/2.

However, if we keep $x - x^3/6$ from $\sin x$ we get
$$ \text{function} \doteq \frac{1}{x} \left( \frac{1}{x} - \frac{1 - x^2/2}{x - x^3/6} \right) = \frac{1}{3} + O(x^2),$$
giving the correct limit 1/3.

When we have cancellations (as in $1/x - \cot x$) we should keep a couple of additional terms just to be sure that all the cancellations are properly accounted for, and so we get a correct constant term.

 

[Orodruin]

This is not quite right, and can sometimes lead to an incorrect answer (as it does in this example).

You misread me. I was referring to the expressions the OP had constructed with a common denominator. Clearly this does lead to the correct limit - as found by the OP.

If you want to be sure about it you should not be throwing terms at all but instead insert O(x^3). This will immediately tell you if you missed a term of a relevant order.