# [SOLVED]Derivative using the limit definition (without using L'Hospital's rule)

#### Prove It

##### Well-known member
MHB Math Helper
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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

#### goody

##### New member
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

##### Well-known member
MHB Math Helper
$$\displaystyle \lim_{\theta \to \pi/2} \tan(\theta)= \infty$$ so $$\displaystyle \lim_{x \to\infty} \arctan(x)=\pi/2$$. That limit certainly does exist!