# Evaluating Limit using Maclaurin series

1. Aug 3, 2010

### 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!

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2. Aug 3, 2010

### 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.

3. Aug 3, 2010

### vela

Staff Emeritus
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.

4. Aug 3, 2010

### Staff: Mentor

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.

5. Aug 4, 2010

### BrownianMan

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.

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6. Aug 4, 2010

### Staff: Mentor

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! + ...)

7. Aug 4, 2010

### vela

Staff Emeritus
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.

8. Aug 5, 2010

Thanks guys!