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

[goody1]

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.

 

[Prove It]

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.

 

[I like Serena]

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.

 

[goody1]

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.

 

[HallsofIvy]

\(\displaystyle \lim_{\theta \to \pi/2} \tan(\theta)= \infty\) so \(\displaystyle \lim_{x \to\infty} \arctan(x)=\pi/2\). That limit certainly does exist!