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- Thread starter goody
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First of all, the denominator is x - x_0, not x - 0.Hello everybody, could you help me with this problem please? I have to find a derivative in x_{0}of this function (without using L'Hospital's rule):

I used the definition , but I don't know what to do next. Thank you.

If you can't use L'Hospital's Rule (which, by the way, is a pointless constraint designed to make life more difficult for the person doing the work), then I'd advise using a series for the arctangent function.

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- #3

- Mar 5, 2012

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So we have:

$$f'(x_0)=\lim_{x\to x_0}\frac{\pi x^2+x\arctan\frac{3\pi}x-f(x_0)}{x-x_0}

=\lim_{x\to 0}\frac{\pi x^2+x\arctan\frac{3\pi}x-f(0)}{x-0}

=\lim_{x\to 0}\Big(\pi x+\arctan\frac{3\pi}x\Big)

$$

Furthermore:

$$\lim_{x\to 0^+}\arctan\frac{3\pi}x = \frac\pi 2$$

$$\lim_{x\to 0^-}\arctan\frac{3\pi}x = -\frac\pi 2$$

So $\lim\limits_{x\to 0}\arctan\frac{3\pi}x$ does not exist, and therefore $f'(x_0)$ does not exist either.

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- #4

- Jan 29, 2012

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