Do we call $x^{\frac{3}{2}}$ discontinous and non differentiable at $ x = 0$

Question

Do we call $x^{\frac{3}{2}}$ discontinous and non differentiable at $ x = 0$ ??

Because the function has the domain as $ x ≥ 0 $ , so we cannot have a LHL for it. So do we call the function continuous or not ?? Is it also differentiable or not as similarly LH Derivative won't exist ??

I'm in highschool, so this might be bit weird , but I got confused , because, many a times, we say (for people who are just getting started ) that a function is continuous if you can make its graph without lifting the pen. So now , to graph it I can draw it without lifting the pen , but the formal definition is that $\lim_{h\rightarrow 0^+} f(x+h) = \lim_{h\rightarrow 0^-} f(x+h) = f(x) $

Answer 1

Formally, limits are not handled by considering a "left" side and a "right" side. The definition of the limit stated in words is something like this:

We say that $\lim\limits_{x\to a}f(x)=L$ iff for any $\epsilon>0$, there is a $\delta>0$ such that we can make $f(x)$ and $L$ within $\epsilon$ of each other for all values of $x$ in the domain of $f$ that are within $\delta$ of $a$.

Symbolically: $\lim\limits_{x\to a}f(x)=L\iff\forall\epsilon>0\exists\delta>0\forall x\in\operatorname{Dom}(f)[0<|x-a|<\delta\Rightarrow|f(x)-L|<\epsilon]$.

See also: $(\epsilon,\delta)$-definition of limit

Provided that the domain of $f$ is a subset of $\mathbb R$ containing both sides of a neighborhood of $a$, this means we must consider the limit from the left as well as the right.

Without the restriction that we only consider the domain of $f$, you might end up considering many different values of $x$. For example, what if $x$ were a complex number? What if it was a 3-dimensional vector? Suddenly, the notion of "left" and "right" become very hard to make use out of. Instead, we make use of the notion of "distance", and from there we define limits.

Continuity is then defined as follows:

$f$ is said to be continuous if, for every $a$ in the domain of $f$, we have $\lim\limits_{x\to a}f(x)=f(a)$, otherwise it is called discontinuous.

And for differentiability:

The derivative of $f$ at $a$ is defined as $\lim\limits_{h\to0}\frac{f(a+h)-f(a)}h$ if it exists, otherwise it is non-differentiable at $a$.

For your given example, all of these conditions are met, so it is both conditinuous and differentiable.

---

And as two more caveats:

• limits to points which cannot be approached from the domain of $f$ are undefined e.g. $\lim\limits_{x\to-1}x^{3/2}$, but if you can approach the point from within the domain of $f$, then you can have a limit, even if $f(a)$ isn't defined.

• for continuity, we do not look at points outside of the domain of $f$ or limits to points outside the domain of $f$. So even if $\lim\limits_{x\to a}f(x)$ exists, but $f(a)$ doesn't, we wouldn't call that a discontinuous function. An example of such is $1/x$.

WolframAlpha seems to give both sides to the $1/x$ being continuous or not. It says:

$\frac1x$ is not continuous on $\mathbb R$.

$\frac1x$ is continuous on its domain.

Though generally we don't consider functions outside of their domain.

Answer 2

That depends on how you define continuity or differentiability. If you define it only on open intervals, then no.

However, there's usually no reason to make such a restriction and we may allow a function to be continuous at a point if the limit exists as one approaches the point in all possible ways (in this case just one), and this limit coincides with the value of the function there. Then in this case the function is both continuous and differentiable at $x=0.$ Indeed, applying the rules formally, you get that $$(x^{3/2})'=\frac 32 \sqrt x,$$ which vanishes when $x=0.$ So there's a finite value for the derivative calculated formally. Also, taking the limit formally yields no problems. Thus we can make life convenient for ourselves.

But again, as a final word, it depends on how much restrictions we placed on ourselves in the first place. It's eventually always up to us.