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

[jokerzz]

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?

 

[Dick]

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?

 

[Unit]

I think your answer is a third too big.

 

[jokerzz]

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

 

[jokerzz]

I think your answer is a third too big.

the answer's 0?

 

[SammyS]

What do you get for the Mclaurin series for the denominator?

 

[Dick]

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).