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TheSodesa

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## Homework Statement

Find [itex]\lim_{x \to 0}\frac{ln(1+x^2)}{1-cos(x)}[/itex] by using series representations. Check using L'Hospitals rule.

## Homework Equations

Taylor polynomial at x=0: [itex]\sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!}(x)^{k} = f(0) + f'(0)(x) + f''(0)x^{2} +...[/itex]

## The Attempt at a Solution

Using L'Hospital's rule gives me a limit of 2. Wolfram Alpha gives the following series expansion for the quotient: http://www.wolframalpha.com/input/?i=series+(ln(1+x^2))/(1-cosx), which obviously evaluates to 2 when [itex]x \to 0[/itex]

What I don't get is how the program calculates the series. Differentiating the quotient given in the problem statement (to find out values of the derivatives to plug into the Taylor-formula) gives very complicated results that either equal zero or are indeterminate when x = 0.

For example, [itex]\frac{d}{dx} \frac{ln(1+x^2)}{1-cos(x)} = \frac{2x(1+x^{2})^{-1}}{1-cos(x)} - \frac{ln(1+x^{2})sin(x)}{(1-cos(x))^{2}}[/itex] where both terms are indeterminate when x = 0. The same thing happens with the second derivative: http://www.wolframalpha.com/input/?i=second+derivative+(ln(1+x^2))/(1-cosx)

So there are indeterminates where things would otherwise seem to evaluate to something that resembles 2, and zeroes everywhere else. It also doesn't help that f(0), the first term in the series, is again indeterminate.

I really have no idea how to approach this problem beyond what I've already tried (which in addition to this approach included plugging in the series for ln(1+x

^{2}) and 1-cos(x) in their respective places in the fraction). Could someone point me in the right direction?

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