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 $\displaystyle x=0$?

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- Mar 21st 2010, 03:58 PMCrazyCat87Differentiability 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 PMchiph588@
$\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 $.