Evaluating Limit using Maclaurin series

In summary, the conversation is discussing how to evaluate a limit by using Maclaurin series and L'Hopital's Rule. The final answer is -1/2, but there was confusion about the steps to get to the answer. It is recommended to write out the series explicitly and factor out the common power of x for easier understanding. The conversation ends with the participants thanking each other for their help.
  • #1
BrownianMan
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0
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!
 

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  • 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
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  • #2
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
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
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
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
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  • #6
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
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
Thanks guys!
 

What is a Maclaurin series?

A Maclaurin series is a special type of power series expansion of a function around a specific point, usually 0. It is named after the Scottish mathematician Colin Maclaurin who first studied and used this method of approximation.

When is it useful to use a Maclaurin series to evaluate a limit?

A Maclaurin series is useful when the function is infinitely differentiable at the point of expansion, and when the limit involves a variable approaching that point. It can also be useful when the function is difficult to evaluate or when the limit is difficult to calculate using other methods.

How do you find the Maclaurin series of a function?

The Maclaurin series of a function can be found by using the Taylor series expansion formula around x = 0. This involves finding the derivatives of the function at x = 0 and plugging them into the formula. Alternatively, the Maclaurin series can also be found by using known Maclaurin series for basic functions and manipulating them using algebraic operations.

What is the general form of a Maclaurin series?

The general form of a Maclaurin series is f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ... + (f^n(0)/n!)x^n, where f(0) is the value of the function at x = 0 and f^(n)(0) is the nth derivative of the function at x = 0.

Can a Maclaurin series be used to evaluate any limit?

No, a Maclaurin series can only be used to evaluate limits that involve a variable approaching the point of expansion, and where the function is infinitely differentiable at that point. It is important to check the conditions for convergence and the validity of the Maclaurin series before using it to evaluate a limit.

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