## Loser66 one year ago If f(x) is differentiable at x0. Prove it continuous at x0 Please, help

1. Loser66

@xapproachesinfinity hey kid, help old man please

2. xapproachesinfinity

hey old man long time

3. xapproachesinfinity

i think theorem state if f is differentiable applies f is continuous so you are proving the theorem

4. xapproachesinfinity

so let see f differentiable at x0 means the limits exist f'(x0)

5. xapproachesinfinity

so $$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=\text{exist}$$

6. xapproachesinfinity

you have to work from here

7. xapproachesinfinity

first let's state the definition of f being cont at x0 that is to say $$\lim_{x\to x_0} f(x)=f(x_0)$$

8. Loser66

Sorry, kid, old man was kicked out of the site.

9. Loser66

But I think you got circulation argument: differentiable --> continuous --

10. Loser66

However, I got it. $$f(x) - f(x_0) = \dfrac{f(x) -f(x_0)}{x-x_0}(x-x_0)$$ $$lim_{x\rightarrow x_0} |f(x) - f(x_0)| = lim_{x\rightarrow x_0}\dfrac{|f(x) -f(x_0)}{|x-x_0|}|x-x_0|$$ since f(x) differentiable at $$x_0$$, the limit exists and as x approaches $$x_0$$ $$|x-x_0|=0$$ That gives us $$lim_{x\rightarrow x_0}|f(x) -f(x_0)|=0$$ or $$lim_{x\rightarrow x_0}f(x) = f(x_0)$$ That shows f(x) continuous at x0

11. xapproachesinfinity

there is no circulation it is one way f different implies f continuous so you start with f different stating the definition limf(x)-f(x0)/x-x0=f'(x0) multilply both sides by (x-x0) we get lim(f(x)-f(x0)=f'(x0)(x-x0) as x tends to x0 the quanties (x-x0) shrinks to zero (limit concept) then we are safe to say lim(f(x)-f(x0)=f'(x0)0=0 therefore lim(f(x))=f(x0) hence f is continuous at x0

12. xapproachesinfinity

a subtlety here is the |x-c| is zero but it is not zero however we write as it is zero