# Differentiability of functions

• Mar 21st 2010, 03:58 PM
CrazyCat87
Differentiability of functions
If the function f is defined as follows:
$\displaystyle f(x)=x^2$ for x rational
$\displaystyle f(x)=0$ for x irrational
How would you show that f is differentiable at $\displaystyle x=0$?
• Mar 21st 2010, 04:10 PM
chiph588@
Quote:

Originally Posted by CrazyCat87
If the function f is defined as follows:
$\displaystyle f(x)=x^2$ for x rational
$\displaystyle f(x)=0$ for x irrational
How would you show that f is differentiable at $\displaystyle x=0$?

$\displaystyle f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We need to show this limit exists when $\displaystyle x=0$.

For this we need to compare $\displaystyle \limsup$ and $\displaystyle \liminf$.

For $\displaystyle x\neq 0$, $\displaystyle x^2 > 0$, so our $\displaystyle \limsup$ will only consider the sequence containing all points of the form $\displaystyle (x,x^2)$.

Thus $\displaystyle \limsup_{h\to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h\to 0}\frac{(0+h)^2-0^2}{h} = \lim_{h\to 0}\frac{h^2}{h} = \lim_{h\to 0}h = 0$.

Clearly our $\displaystyle \liminf$ will only consider the sequence containing all points of the form $\displaystyle (x,0)$.

Thus $\displaystyle \liminf_{h\to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h\to 0}\frac{0}{h} = 0$.

Since $\displaystyle \limsup=\liminf$, $\displaystyle \lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$ exists and equals $\displaystyle 0$.
Therefore $\displaystyle f'(0)=0$.