Differentiability

[Imum Coeli]

Just wondering of someone could check this. My answer seems awfully short.

Q: Show that \(\displaystyle f(x) = |x|\) is not differentiable at \(\displaystyle x=0\)

A: If \(\displaystyle f(x) = |x|\) is not diff'able at \(\displaystyle x=0 \) then \(\displaystyle \exists \; \delta >0 \) such that \(\displaystyle 0<|x|<\delta \implies |\frac{|x|}{x}|>\epsilon \)
Suppose that \(\displaystyle \epsilon = 1/2\)
Now if \(\displaystyle 0<|x|<\delta \) then \(\displaystyle |\frac{|x|}{x}|=1 \;\forall x\)
Since \(\displaystyle 0<|x|<\delta \implies|\frac{|x|}{x}|=1 > \epsilon\) then \(\displaystyle f(x) = |x|\) is not differentiable at \(\displaystyle x=0\)

Thanks.

**EDIT
Well I noticed a large oversight. I assumed that the limit exists and is equal to 0 when it should be some real L.

 

[Ishuda]

Since f(0) is zero, if f were differentiable at x=0 then
\(\displaystyle \lim_{x\to0}\frac{f(x)} {x}\)
would exist and be the same no matter how you got there. Consider what happens as you let x go to zero from the right [f(x)=x] and then what happens when you let x go to zero from the left[f(x)=-x].

 

[Imum Coeli]

Since f(0) is zero, if f were differentiable at x=0 then
\(\displaystyle \lim_{x\to0}\frac{f(x)} {x}\)
would exist and be the same no matter how you got there. Consider what happens as you let x go to zero from the right [f(x)=x] and then what happens when you let x go to zero from the left[f(x)=-x].

Well I'm not allowed to 'state facts about one-sided limits' so the best I can do is the following:

Suppose that \(\displaystyle \lim_{x \to 0} \frac{|x|}{x}=L\)

Let \(\displaystyle \epsilon=1 \) then \(\displaystyle \exists \, \delta>0\,:\,|\frac{|x|}{x}-L|<1\; \forall \; 0< |x-0|<\delta\)

Pick \(\displaystyle x= \frac{\delta}{\delta+1} \) and \(\displaystyle x= \frac{-\delta}{\delta+1} \),

then \(\displaystyle 0<|\frac{\delta}{\delta+1}-0|<\delta \) and \(\displaystyle 0<|\frac{-\delta}{\delta+1}-0|<\delta \)

so \(\displaystyle |\frac{|x|}{x}-L|<\epsilon \implies |1-L|<\epsilon\) and \(\displaystyle |-1-L|<\epsilon\) respectively

but \(\displaystyle |1-L|<\epsilon \iff -1<1-L<1 \iff 0<L<2\)

and \(\displaystyle |1-L|<\epsilon \iff -1<-1-L<1 \iff -2<L<0\)

which is impossible because limits are unique.

\(\displaystyle \therefore \lim_{x \to 0} \frac{|x|}{x}\neq L\)

and hence \(\displaystyle f(x) = |x|\) is not differentiable at \(\displaystyle x=0\)

 

[Imum Coeli]

I'm not sure if this needs it's own thread because it's technically a different question but...

Let's say I want to prove that \(\displaystyle \lim_{x \to 0} \frac{|x|^3}{x} \)exists


Is it enough to suppose that \(\displaystyle \lim_{x \to 0} \frac{|x|^3}{x}=0 \)

then given \(\displaystyle \epsilon >0\), let \(\displaystyle \delta := \frac{\epsilon}{\epsilon+1}\) (note \(\displaystyle \delta<1\))

so if \(\displaystyle 0<|x|<\delta\) then \(\displaystyle |\frac{|x|^3}{x}|<|\pm(\frac{\epsilon}{\epsilon+1})^2|<\epsilon\)

Therefore \(\displaystyle \lim_{x \to 0} \frac{|x|^3}{x} \) does exist and is equal to zero.