Find the limit using taylor series

Tags:
1. Jan 2, 2017

doktorwho

1. The problem statement, all variables and given/known data
Using the taylor series at point $(x=0)$ also known as the meclaurin series find the limit of the expression:
$$L=\lim_{x \rightarrow 0} \frac{1}{x}\left(\frac{1}{x}-\frac{cosx}{sinx}\right)$$
2. Relevant equations
3. The attempt at a solution

$L=\lim_{x \rightarrow 0} \frac{1}{x}\left(\frac{sinx-xcosx}{xsinx}\right)$
$L=\lim_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{sinx-xcosx}{sinx}\right)$
I need help to continue. This is the first problem im 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?

2. Jan 2, 2017

Orodruin

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

3. Jan 2, 2017

Staff: Mentor

Shorter is directly the series for cotangent.

4. Jan 2, 2017

vela

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

5. Jan 2, 2017

vela

Staff Emeritus
Yeah, but who remembers that?

6. Jan 2, 2017

Staff: Mentor

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

7. Jan 2, 2017

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?

8. Jan 2, 2017

Staff: Mentor

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.)

9. Jan 2, 2017

doktorwho

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?

10. Jan 2, 2017

Staff: Mentor

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".)

11. Jan 2, 2017

Ray Vickson

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.

Last edited: Jan 3, 2017
12. Jan 2, 2017

Orodruin

Staff Emeritus
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.