Limit of a Rational Function without L'Hospital's Rule

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Homework Help Overview

The discussion revolves around evaluating the limit of a rational function as x approaches 0, specifically the expression (e^xsin(x)-x-x^2)/(x^3 +xln(1-x)). The subject area includes calculus and series expansions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the Maclaurin series for e^x and sin(x) to evaluate the limit. There are questions about the treatment of terms in the denominator, particularly regarding the presence of an x^2 term and its implications for the limit calculation.

Discussion Status

The discussion is ongoing, with participants providing different interpretations of the limit and questioning each other's reasoning. Some guidance has been offered regarding the series expansions, but no consensus has been reached on the correct limit value.

Contextual Notes

There is some confusion regarding the terms in the denominator, particularly whether an x^2 term is present, which affects the limit evaluation. Participants are encouraged to clarify their steps and reasoning further.

jokerzz
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Lim x-> 0 (e^xsin(x)-x-x^2)/(x^3 +xln(1-x))


I first put the Mclaurin series of e^x and sinx and then took x^3 common and then put the limit. My answer comes out to be 1/3. Can anyone confirm whether I've done it right?
 
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jokerzz said:
Lim x-> 0 (e^xsin(x)-x-x^2)/(x^3 +xln(1-x))


I first put the Mclaurin series of e^x and sinx and then took x^3 common and then put the limit. My answer comes out to be 1/3. Can anyone confirm whether I've done it right?

It would help if you showed a little more of the details, but there is an x^2 term in the denominator as well, isn't there? What did you do with that?
 
I think your answer is a third too big.
 
Dick said:
It would help if you showed a little more of the details, but there is an x^2 term in the denominator as well, isn't there? What did you do with that?

no there's no x^2. When you put x=0 at the end after substituting the mclaurin series of e^x and sinx, everything becomes 0, except -1/3! + 1/2 which is equal to 1/3
 
Unit said:
I think your answer is a third too big.


the answer's 0?
 
What do you get for the Mclaurin series for the denominator?
 
jokerzz said:
no there's no x^2. When you put x=0 at the end after substituting the mclaurin series of e^x and sinx, everything becomes 0, except -1/3! + 1/2 which is equal to 1/3

There's an x^2 in the expansion of x*log(1-x).
 

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