Find the limit using taylor series

In summary: I know, it's the same as keeping a few terms - but it's a bit shorter.)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. (I know, it's the same as keeping a few terms - but it's a bit shorter.)I didn't misread you. The OP was asking for help, and I was addressing the question of when one might keep only a few terms. I gave an example
  • #1
doktorwho
181
6

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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  • #2
doktorwho said:
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.
 
  • #3
Shorter is directly the series for cotangent.
 
  • #4
You may also find the series ##\frac{1}{1-z} = 1+z+z^2+z^3+\cdots## useful.
 
  • #5
fresh_42 said:
Shorter is directly the series for cotangent.
Yeah, but who remembers that? :wink:
 
  • #6
vela said:
Yeah, but who remembers that? :wink:
Wiki does. The advantage is, you don't even need a pencil.:smile:
 
  • #7
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?
 
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  • #8
doktorwho said:
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.)
 
  • #9
fresh_42 said:
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?
 
  • #10
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".)
 
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  • #11
doktorwho said:

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 said:
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:
  • #12
Ray Vickson said:
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.
 

Related to Find the limit using taylor series

1. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a specific point. It is used to approximate the value of a function at a given point.

2. How do you find the limit using Taylor series?

To find the limit of a function using Taylor series, you first need to find the Taylor series representation of the function. Then, you can take the limit of the series as the number of terms approaches infinity, which will give you an approximation of the limit of the original function.

3. What is the benefit of using Taylor series to find a limit?

The benefit of using Taylor series to find a limit is that it can often provide a more accurate approximation than traditional methods, such as L'Hopital's rule. It can also be used for functions that are not easily differentiable or have complex derivatives.

4. Are there any limitations to using Taylor series to find a limit?

Yes, there are limitations to using Taylor series to find a limit. The series may not converge for all values of the input, and the approximation may not be accurate for functions with non-continuous or non-differentiable points. Additionally, higher order terms in the series may be difficult to calculate, making it impractical for some functions.

5. Can Taylor series be used to find the limit of any function?

No, Taylor series can only be used to find the limit of functions that can be represented by a convergent series. It is not applicable to all types of functions, such as functions with essential singularities or functions that do not have a Taylor series representation.

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