Evaluating Limit using Maclaurin series

[BrownianMan]

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!

 

[Dick]

Substitute the series. What's the lowest power of x that doesn't cancel in numerator and denominator? Cancel it as a common factor.

 

[vela]

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.

 

[Mark44]

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.

 

[BrownianMan]

Yes, the answer is -1/2!

However, I'm still having some problems.

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

 

[Mark44]

Don't write sin(x) and 2e^x as summations. Instead, write sin(x) as x - x³/3! + x⁵/5! -+ ...
and write 2e^x as 2(1 + x + x²/2! + x³/3! + ...)

 

[vela]

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.

 

[BrownianMan]

Thanks guys!