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: $$\forall\epsilon>0, \exists\delta, 0<|x-a|<\delta \implies |[f(x) - f(a)]-0|<\epsilon \implies |f(x) - f(a)|<\epsilon$$
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. =)