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Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Please, see if I made some mistake in the proof below. I mention some theorems in the proof:

The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.

(1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$.

(2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$.

(3) We'll do this by the Precise Limit Definion: given $ \epsilon>0, \exists\delta|0<|x-x_0|<\delta$, then $0<|f(x)-f(x_0)|<\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at $x=x_0$. The end.

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    $\begingroup$ Where is your proof? You announce that you are going to do something but I fail to see that you do it. $\endgroup$ – Did May 18 '13 at 20:18
  • $\begingroup$ A minor nitpick: It's $f$ that's differentiable/continuous at $x_0$. When we write $f(x)$ we're talking about the value of $f$ at $x$, a real number (in this case). Now, old habits die hard, and it's still common for analysts to talk about "the function $f(x)$" (especially when talking about complex functions), but technically it's wrong. $\endgroup$ – kahen May 18 '13 at 20:26
  • $\begingroup$ @kahen if you call your function $f(x)$ then $f(x)$ will be differentiable in a point and the function value will be $f(x)(x)$. (Just kidding) $\endgroup$ – Dominic Michaelis May 18 '13 at 20:28
  • $\begingroup$ @Did, I enumerated the statements. Briefly, I tried to proof by this implications: $f(x)$ is differentiable $\overset{(1)}{\Rightarrow} f'(x)$ exists $\overset{(2)}{\Rightarrow} \lim_{x\to x_0} f(x)$ also exists $\overset{(3)}{\Rightarrow} \lim_{x\to x_0} f(x) = f(x_0)$. As others said, maybe my error was was not developing $(3)$ well. $\endgroup$ – srodriguex May 19 '13 at 1:09
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$$\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0}\left(f(x_0)+(x-x_0)\cdot\dfrac {f(x)-f(x_0)}{x-x_0}\right)=$$ $$\lim\limits_{x\to x_0}f(x_0)+\lim\limits_{x\to x_0}(x-x_0)\cdot\lim\limits_{x\to x_0}\dfrac {f(x)-f(x_0)}{x-x_0}=$$ $$f(x_0)+0\cdot f'(x_0)=f(x_0)$$

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  • $\begingroup$ by the way a very neat solution :) $\endgroup$ – Dominic Michaelis May 19 '13 at 15:15
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You surely mess up things in the proof. You say we can make $f(x)$ as close to $f(x_0)$ as one whishes. But this is exactly what you are trying to prove.

The more useful definition of differentiability is, that a function is differentiable if $f(x)=f(x_0) + a \cdot (x-x_0) + r(x)$ with $$\lim_{x\to x_0} \frac{r(x)}{x-x_0} = 0 $$ so $\lim_{x\to x_0} r(x)$ will surely be $0$. Hence \begin{align*} |f(x)-f(x_0)|&= | f(x_0) + a \cdot (x-x_0) + r(x) -f(x_0)|\\ &\leq |a| \cdot |x-x_0|+ |r(x)| \end{align*} Now you have a sum of 2 terms which go to zero as $x$ goes to $x_0$.

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  • $\begingroup$ I couldn't use the definition you mentioned, but I liked answer. $\endgroup$ – srodriguex May 19 '13 at 1:12
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Here, a solution I've found at "Introduction to Analysis" byt Arthur Mattuck:

$$\lim_{x\to x_0} (f(x)-f(x_0)) = \lim_{x\to x_0} (\dfrac{f(x)-f(x_0)}{x-x_0})(x-x_0)\\ = \lim_{x\to x_0} (\dfrac{f(x)-f(x_0)}{x-x_0}) \cdot \lim_{x\to x_0} (x-x_0)\\ = f'(x)\cdot 0 \\ =0$$

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