MHB Derivative using the limit definition (without using L'Hospital's rule)

AI Thread Summary
The discussion focuses on finding the derivative of a function at a point x0 without using L'Hospital's rule. Participants clarify that the denominator should be x - x0 and suggest using a series for the arctangent function. The limit approaches show that the derivative does not exist due to the differing limits from the left and right of zero for the arctangent function. The conversation highlights the challenge of solving the problem without L'Hospital's rule, emphasizing the importance of understanding limits. Ultimately, the solution reveals that the limit of the derivative does not exist, leading to a deeper understanding of the function's behavior.
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Hello everybody, could you help me with this problem please? I have to find a derivative in x0 of this function (without using L'Hospital's rule):
View attachment 9694

I used the definition View attachment 9695, but I don't know what to do next. Thank you.
 

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goody said:
Hello everybody, could you help me with this problem please? I have to find a derivative in x0 of this function (without using L'Hospital's rule):I used the definition , but I don't know what to do next. Thank you.

First of all, the denominator is x - x_0, not x - 0.

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.
 
Note that $f(x_0)=f(0)$ has been defined as $0$.

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.
 
I know using L'Hospital's rule would be easy way to solve it but we haven't learned it yet so we're forced to find another ways.

Anyways, thank you so much Klaas van Aarsen, now when you showed me it looks so simple.
 
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