Differentiability implies continuity proof (delta epsilon)

[Nan1teZ]

1. The problem statement.

Give a complete and accurate \delta - \epsilon proof of the thereom: If f is differentiable at a, then f is continuous at a.

2. The attempt at a solution

Known:
\forall\epsilon>0, \exists\delta>0, \forall x, |x-a|<\delta \implies \left|\frac{f(x) - f(a)}{x-a} - f'(a)\right|<\epsilon

Want to show:

\forall\epsilon>0, \exists\delta>0, \forall x, |x-a|<\delta \implies |f(x) - f(a)|<\epsilon

So I start with the known info and cross multiply \left|\frac{f(x) - f(a)}{x-a} - f'(a)\right| to get \left|\frac{f(x) - f(a) - (x-a)f'(a)}{x-a}\right| which doesn't really help me in completing the proof, especially since x-a is in the denominator. =[

And is my known and want to show info correct?

 

[konthelion]

The known and show info look correct. As for the proof, hint: f(x)-f(a)= \frac{f(x)-f(a)}{x-a}*(x-a) and \lim_{x \to a}(x-a)=0

In the end you can should be able to prove that \lim_{x \to a}f(x)-f(a)=0 i.e. |f(x) - f(a)|<\epsilon

 

[Nan1teZ]

As for the proof, hint: f(x)-f(a)= \frac{f(x)-f(a)}{x-a}*(x-a) and \lim_{x \to a}(x-a)=0

Does that mean we can assume \frac{f(x)-f(a)}{x-a}*(x-a) = 0?

 

[konthelion]

Let me rephrase that.

From what you know:
\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=f'(a) which when written in \epsilon - \delta definition,\left|\frac{f(x) - f(a)}{x-a} - f'(a) \right|<\epsilon

Also, since f is differentiable at a, then \lim_{x \to a}(x-a)=0

You are trying to show that \lim_{x \to a}f(x)=f(a) , by the definition of continuity at a. Which is also written in \epsilon - \delta definition as
|f(x) - f(a)|<\epsilon

From the hint I gave you, \lim_{x \to a}f(x)-f(a)=blank

 

[Nan1teZ]

Let me rephrase that.

From what you know:
\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=f'(a) which when written in \epsilon - \delta definition,\left|\frac{f(x) - f(a)}{x-a} - f'(a) \right|<\epsilon

Also, since f is differentiable at a, then \lim_{x \to a}(x-a)=0

You are trying to show that \lim_{x \to a}f(x)=f(a) , by the definition of continuity at a. Which is also written in \epsilon - \delta definition as
|f(x) - f(a)|<\epsilon

From the hint I gave you, \lim_{x \to a}f(x)-f(a)=blank

Okay so tell me if this is right:

\left|f(x)-f(a)\right| = \left|\frac{f(x)-f(a)}{x-a}(x-a)\right| = \left|\frac{f(x)-f(a)}{x-a}(0)\right| = 0 < \epsilon


Since \lim_{x \to a}\frac{f(x)-f(a)}{x-a}*(x-a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}*\lim_{x \to a}(x-a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}*0 = 0

 

[konthelion]

Since \lim_{x \to a}\frac{f(x)-f(a)}{x-a}*(x-a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}*\lim_{x \to a}(x-a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}*0 = 0

Yes, this part is correct. You can further simplify that into:

\lim_{x \to a}\frac{f(x)-f(a)}{x-a}*0 = f'(a)*0 = 0

Then, you can just say that: \forall\epsilon&gt;0, \exists\delta, 0&lt;|x-a|&lt;\delta \implies |[f(x) - f(a)]-0|&lt;\epsilon \implies |f(x) - f(a)|&lt;\epsilon<br />
since you've shown that \lim_{x \to a}[ f(x)-f(a)]=0

You can't say that (x-a)=0 since the limit \lim_{x \to a}(x-a)=0

 

[Nan1teZ]

Okay thanks a lot. =)