# 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?

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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.

Mentor
2022 Award
Shorter is directly the series for cotangent.

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

Staff Emeritus
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Shorter is directly the series for cotangent.
Yeah, but who remembers that? Mentor
2022 Award
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?

• Logical Dog
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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?

Mentor
2022 Award
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".)

• doktorwho
Homework Helper
Dearly Missed

## 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.

Last edited:
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