Evaluating Limit using Maclaurin series

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Homework Help Overview

The discussion revolves around evaluating a limit using Maclaurin series, specifically involving the functions sin(x) and 2e^x. Participants are exploring the steps necessary to manipulate the limit into a form suitable for evaluation at x=0.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss substituting Maclaurin series for sin(x) and 2e^x, questioning how to identify the lowest power of x that remains after cancellation in the limit. There are also inquiries about the transition between different forms of the limit.

Discussion Status

The conversation includes differing opinions on the value of the limit, with some participants suggesting alternative methods such as L'Hopital's Rule. Guidance has been offered regarding the explicit representation of series terms to facilitate understanding.

Contextual Notes

There is mention of a textbook example and a specific final answer provided, which may influence participants' expectations and understanding of the problem. The discussion reflects uncertainty about the necessity of using Maclaurin series for the evaluation.

BrownianMan
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gif.latex?\lim_{x&space;\to&space;0}\frac{\sin&space;x-x}{2e^x-2-2x-x^2}.gif


This is an example given in my textbook. The final answer is 1/6. I know that sinx and 2e^x have to be replaced with their corresponding Maclaurin series. However, I'm having trouble understanding the steps they took to get the limit in a form in which it could be evaluated by substituting x=0.

If anyone could help, I'd appreciate it!
 

Attachments

  • gif.latex?\lim_{x&space;\to&space;0}\frac{\sin&space;x-x}{2e^x-2-2x-x^2}.gif
    gif.latex?\lim_{x&space;\to&space;0}\frac{\sin&space;x-x}{2e^x-2-2x-x^2}.gif
    788 bytes · Views: 716
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Substitute the series. What's the lowest power of x that doesn't cancel in numerator and denominator? Cancel it as a common factor.
 
The limit is actually equal to -1/2, not 1/6, so if you think you don't understand it because you're getting the "wrong" answer, that might be why.
 
I concur with vela on the value of the limit.

If you don't actually have to use Maclaurin series to evaluate this limit, it can be found by applying L'Hopital's Rule a few times.
 
Yes, the answer is -1/2!

However, I'm still having some problems.

2k+1&space;\right&space;)!}}{\frac{2x^3}{6}+x^4\sum_{k=4}^{\infty}\frac{2x^{k-4}}{k!}}.gif


I'm not understanding how they went from the limit on the left to the one on the right.
 

Attachments

  • 2k+1&space;\right&space;)!}}{\frac{2x^3}{6}+x^4\sum_{k=4}^{\infty}\frac{2x^{k-4}}{k!}}.gif
    2k+1&space;\right&space;)!}}{\frac{2x^3}{6}+x^4\sum_{k=4}^{\infty}\frac{2x^{k-4}}{k!}}.gif
    2.6 KB · Views: 649
Don't write sin(x) and 2ex as summations. Instead, write sin(x) as x - x3/3! + x5/5! -+ ...
and write 2ex as 2(1 + x + x2/2! + x3/3! + ...)
 
They just wrote the first term of the series explicitly and then factored out the common power of x from the remaining terms. It's easier to see what's going on in this type of problems if you just write out the series like Mark44 suggested.
 
Thanks guys!
 

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