Question

If f(x) is differentiable at x0. Prove it continuous at x0
Please, help

Answer 1

@xapproachesinfinity hey kid, help old man please

Answer 2

hey old man long time

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

you have to work from here

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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