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I have always treated them as the same thing. But recently, some people have told me that the two terms are different. So now I am wondering,

What is the difference between "differentiable" and "continuous"?

I just don't want to say the wrong thing. For example, I don't want to say, "$\frac{x^2}{x^4-2x^3}$ is not differentiable at $x=0$" when really, it should be "discontinuous". Please help

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    $\begingroup$ Do you know the definitions of continuity at $x$ and of differentiability at $x$? $\endgroup$
    – Git Gud
    Mar 23 '14 at 18:26
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Differentiability is a stronger condition than continuity. If $f$ is differentiable at $x=a$, then $f$ is continuous at $x=a$ as well. But the reverse need not hold.

Continuity of $f$ at $x=a$ requires only that $f(x)-f(a)$ converges to zero as $x\rightarrow a$.

For differentiability, that difference is required to converge even after being divided by $x-a$. In other words, $\dfrac{f(x)-f(a)}{x-a}$ must converge as $x\rightarrow a$.

Not that if that fraction does converge, the numerator necessarily converges to zero, implying continuity as I mentioned in the first paragraph.

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  • $\begingroup$ I think that the last two sentences in this answer are not accurate. A function $f(x)$ is differentiable at $x=a$ if the derivative exists there, but it doesn't have to be zero. Thus, your method of showing that continuity follows differentiability by having a zero numerator would be flawed. A more accurate way is to say that: $\lim\limits_{x \to a}[f(x)-f(a)]=\lim\limits_{x \to a}[\frac{f(x)-f(a)}{x-a}]\lim\limits_{x \to a}(x-a)=f'(a)\lim\limits_{x \to a}(x-a)$ which, when $f'(a)=c$ exists, equals: $c \lim\limits_{x \to a}(x-a)=c (0)=0$ and therefore differentiability implies continuity. $\endgroup$
    – user135626
    Sep 14 '16 at 0:28
  • $\begingroup$ @user135626: What I wrote is correct. You are misreading it. I'm not saying the derivative is zero, I'm saying that if the derivative exists, the numerator of the difference quotient necessarily converges to zero (not that the difference quotient itself must). That's just because of the fact that the denominator does, since otherwise the difference quotient itself could not converge at all. Also, your reasoning is flawed: one cannot in general write that $\lim AB=\lim A\lim B$. $\endgroup$
    – MPW
    Sep 14 '16 at 7:12
  • $\begingroup$ I see what you mean now, and I agree (sorry for misreading your answer). However, I don't think that the method I wrote is flawed, because Limit Laws state that, if $\lim A$ and $\lim B$ exist, then $\lim(AB)=\lim A\lim B$. Indeed, the same method can be found in many classic texts. For example, see p.75 in Simmon's classic book, Calculus with Analytic Geometry. $\endgroup$
    – user135626
    Sep 14 '16 at 21:50
  • $\begingroup$ @user135626: Yes, I know that. Perhaps "flawed" was too harsh. I just meant that you need to be careful when breaking up a limit like that. You've given proper justification in your response though. Good. $\endgroup$
    – MPW
    Sep 15 '16 at 12:43
  • $\begingroup$ I must say that your original answer to the question is actually elegantly pithy, when carefully read. Many thanks. $\endgroup$
    – user135626
    Sep 15 '16 at 17:34
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Differentiability means that the function has a derivative at a point.

Continuity means that the limit from both sides of a value is equal to the function's value at that point.

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The typical example is $f(x)=|x|$. It is continuous for all $x$, but has a corner at $x=0$ and is not differentiable there.

Your example is not defined at $x=0$. This is stronger than not continuous, which in turn is stronger than not differentiable.

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  • $\begingroup$ Kind of off-topic but if infinitely differentiable implies differentiable which implies continuous that implies defined is there something stronger than infinitely differentiable? $\endgroup$
    – ruler501
    Mar 23 '14 at 18:38
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    $\begingroup$ @ruler501: "analytic" is stronger, en.wikipedia.org/wiki/Analytic_function $\endgroup$
    – M. E.
    Mar 23 '14 at 18:44
  • $\begingroup$ Ah. I forgot about that condition. $\endgroup$
    – ruler501
    Mar 23 '14 at 18:45

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