Loser66
  • Loser66
If f(x) is differentiable at x0. Prove it continuous at x0 Please, help
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
@xapproachesinfinity hey kid, help old man please
xapproachesinfinity
  • xapproachesinfinity
hey old man long time
xapproachesinfinity
  • xapproachesinfinity
i think theorem state if f is differentiable applies f is continuous so you are proving the theorem

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

xapproachesinfinity
  • xapproachesinfinity
so let see f differentiable at x0 means the limits exist f'(x0)
xapproachesinfinity
  • xapproachesinfinity
so \(\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=\text{exist}\)
xapproachesinfinity
  • xapproachesinfinity
you have to work from here
xapproachesinfinity
  • xapproachesinfinity
first let's state the definition of f being cont at x0 that is to say \(\lim_{x\to x_0} f(x)=f(x_0)\)
Loser66
  • Loser66
Sorry, kid, old man was kicked out of the site.
Loser66
  • Loser66
But I think you got circulation argument: differentiable --> continuous --
Loser66
  • 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
xapproachesinfinity
  • 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
xapproachesinfinity
  • xapproachesinfinity
a subtlety here is the |x-c| is zero but it is not zero however we write as it is zero

Looking for something else?

Not the answer you are looking for? Search for more explanations.