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

goody

New member
Mar 31, 2020
14
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):
limm.png

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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
Mar 5, 2012
8,713
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
Mar 31, 2020
14
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
Jan 29, 2012
1,151
\(\displaystyle \lim_{\theta \to \pi/2} \tan(\theta)= \infty\) so \(\displaystyle \lim_{x \to\infty} \arctan(x)=\pi/2\). That limit certainly does exist!