Continuity and limit of square root function: Does limit at 0 not exist?

Delta

New member
Joined
Jun 10, 2018
Messages
9
Hi.

I have 2 questions, first:

Say that there's f(x) = sqrt(x), I want to find the limit of the function at x = 0, so I have to check the right and left sided limits:

lim x->0+ sqrt(x) = sqrt(0) = 0
lim x->0- sqrt(x) = ?? (function isn't defined on a small interval to the left of zero)

thus, this limit doesn't exist right?


Second question:

It is said that if f(x) is continuous at x = a, and a is inside the interval f(x) >= 0, then g(x) = sqrt(f(x)) would be continuous at a
, what if a is at the edge of g(x) ? i mean it's a zero of the function g(x) (like 0 is the zero for sqrt(x)), wouldn't this function be discontinuous at x = a?

Thanks.
 
Hi.

I have 2 questions, first:

Say that there's f(x) = sqrt(x), I want to find the limit of the function at x = 0, so I have to check the right and left sided limits:

lim x->0+ sqrt(x) = sqrt(0) = 0
lim x->0- sqrt(x) = ?? (function isn't defined on a small interval to the left of zero)

thus, this limit doesn't exist right?


Second question:

It is said that if f(x) is continuous at x = a, and a is inside the interval f(x) >= 0, then g(x) = sqrt(f(x)) would be continuous at a
, what if a is at the edge of g(x) ? i mean it's a zero of the function g(x) (like 0 is the zero for sqrt(x)), wouldn't this function be discontinuous at x = a?

Thanks.
The first problem is tricky. IMO it depends on how you think about it. I would say that the function has a limit at x=0, because the function is only defined for non-negative numbers and you did get a right hand limit. Now if I were to be completely formal and just use what you gave us then I would do what you did and after observing that the left hand limit and right hand limit did not match, then I would say that the limit does not exist.
 
Yup, that makes sense right? Some online calculators apply what I said and some don't.
 
Anymore ideas or opinions?

It is typically stated that a function has no limit at a boundary of its (closed) domain, because one of the one-sided limits does not exist. If your textbook or other context says this, then go with it.

It is also possible under some circumstances to extend the concept of limit to cover this case, particularly if you think of the domain as all there is, so that there are no points outside of it. If your textbook allows for this, then go with it.

What you have done, however, is to refer to unidentified sources ("it is said that") with no context. Can you identify them, and quote their definitions? Whether a limit exists depends on how you define limits.
 
True. My book states that the limit at the edge of a square root function doesn't exist.

What you have done, however, is to refer to unidentified sources ("it is said that") with no context. Can you identify them, and quote their definitions? Whether a limit exists depends on how you define limits.

You mean the second question? Ah well, my book states that.
 
Second question:

It is said that if f(x) is continuous at x = a, and a is inside the interval f(x) >= 0, then g(x) = sqrt(f(x)) would be continuous at a
, what if a is at the edge of g(x) ? i mean it's a zero of the function g(x) (like 0 is the zero for sqrt(x)), wouldn't this function be discontinuous at x = a?

True. My book states that the limit at the edge of a square root function doesn't exist.

You mean the second question? Ah well, my book states that.

So, the book's definition, like most, says that the square root function has no limit at 0; but it also says that g(x) = √(f(x)) is continuous on the entire domain of f. In particular, to take a very simple example, if f(x) = x, then f is continuous at 0, and 0 is in the interval where f(x) >= 0, so g(x) = √x is continuous at 0. That seems inconsistent, right?

We have a couple more questions to ask of the book: First, if you quoted exactly, does "inside the interval" mean the same thing as "in the interval", or might it refer to the interior of that interval, namely f(x) > 0? Second, how does it define continuity at the endpoint of a domain? Here, as with limits, there are ways to define continuity so that a two-sided limit is not required. Can you quote their definition? And if you didn't quote exactly what the book says about your second question, can you do that?
 
So, the book's definition, like most, says that the square root function has no limit at 0; but it also says that g(x) = √(f(x)) is continuous on the entire domain of f. In particular, to take a very simple example, if f(x) = x, then f is continuous at 0, and 0 is in the interval where f(x) >= 0, so g(x) = √x is continuous at 0. That seems inconsistent, right?

We have a couple more questions to ask of the book: First, if you quoted exactly, does "inside the interval" mean the same thing as "in the interval", or might it refer to the interior of that interval, namely f(x) > 0?

Here's the literal quotation translated from Arabic into English concerning the square root theorm.

"If f was a function that's continuous at x = a, f(x) >= 0, in an open interval that contains a, then g(x) = √f(x) is continuous at x = a"

What do you understand from that? it mentions a closed interval f(x) >= 0 then states "in an open interval".. Unless it means something else

Second, how does it define continuity at the endpoint of a domain? Here, as with limits, there are ways to define continuity so that a two-sided limit is not required. Can you quote their definition? And if you didn't quote exactly what the book says about your second question, can you do that?


As far as that is concerned, it states literally that:

"If f is a function that's defined on the interval [a,b], then:

1) f is continuous at x = a from the right, if lim f(x) x -> a+ = f(a)
2)
f is continuous at x = b from the left, if lim f(x) x -> b- = f(b)
3) f is continuous on (a,b) if it was continuous at every x such that x is in (a,b)
4) f is continuous on [a,b] if it was continuous on (a,b) and continuous at a from the right and continuous at b from the left"



P.S: Sorry for late reply
 
Here's the literal quotation translated from Arabic into English concerning the square root theorm.

"If f was a function that's continuous at x = a, f(x) >= 0, in an open interval that contains a, then g(x) = √f(x) is continuous at x = a"

What do you understand from that? it mentions a closed interval f(x) >= 0 then states "in an open interval".. Unless it means something else

The open interval is a set of values of x, not of f(x). The comma you put before "in an open interval" somewhat obscures the meaning.

For example, if f is continuous at x = 1, and f(x) is non-negative in the open interval 0 < x < 2, then g(x) = √f(x) is continuous at x = 1.

As an example, suppose that f(x) = |x-1|, which is continuous at x = 1, and is non-negative everywhere. Then g(x) = √|x-1|; if you graph this, you can see that it is continuous at x = 1, and if you consider limits, as x approaches 1, |x-1| approaches 0, and √|x-1| approaches 0 because, regardless of the direction from which x approaches 1, |x-1| is approaching 0 from the right. You could do the same with f(x) = (x-1)^2.

The theorem would not apply to my previous example with f(x) = x, which is not non-negative in a open interval around x = 0.

I don't think you mentioned the open interval before.

As far as that is concerned, it states literally that:

"If f is a function that's defined on the interval [a,b], then:

1) f is continuous at x = a from the right, if lim f(x) x -> a+ = f(a)
2) f is continuous at x = b from the left, if lim f(x) x -> b- = f(b)
3) f is continuous on (a,b) if it was continuous at every x such that x is in (a,b)
4) f is continuous on [a,b] if it was continuous on (a,b) and continuous at a from the right and continuous at b from the left"


This is the extension I referred to, where we can say that the function is continuous on a closed interval if is it "as continuous as possible" everywhere: it can only be continuous from the left if it is not defined to the right, so we are accepting that as sufficient.

That is very helpful. We can now say that f(x) = √x is continuous on its domain, [0,infty). (They didn't explicitly mention half-closed intervals, but I'm sure this is what they intend!)

In mathematics, every detail counts; this is why stating a problem in full, and stating theorems including all their conditions, is essential.
 
Now I do get why they mentioned "open interval"... And yeah so when we say that √x is continuous we actually mean that it's continuous on its domain only [0,+inf), but if we were to examine its continuity over real numbers set, it wouldn't be continuous in all R.

And yeah actually that comma is there in the theorem context.

Two last things that i still have questions for; they said "f(x)
>= 0" and you said "f(x) is non-negative in the open interval 0 < x < 2" Are they mistaken or something? Like, shouldn't they have stated "f(x) > 0" rather?

Secondly, what if f(x) is continuous at b and b is the end point of g(x) =
√f(x) such that g(x)'s domain is [b, +inf), we can say that g(x) is continuous from the right at x = b, correct? But not fully continuous (by fully continuous i mean the global limit exists; on the right and left).



 
Now I do get why they mentioned "open interval"... And yeah so when we say that √x is continuous we actually mean that it's continuous on its domain only [0,+inf), but if we were to examine its continuity over real numbers set, it wouldn't be continuous in all R.

Two last things that i still have questions for; they said "f(x)
>= 0" and you said "f(x) is non-negative in the open interval 0 < x < 2" Are they mistaken or something? Like, shouldn't they have stated "f(x) > 0" rather?


No; f(x) ≥ 0 means f(x) is non-negative (not negative: either positive or zero). If they said f(x) > 0, that would mean f(x) must be positive (not zero), and that is not what they want to say. L
ook again at my example: |x-1| is either zero or positive in the open interval 0<x<2.

Why do you think
f(x) > 0 would be correct? Are you confusing what is true of f(x) with what is true of x?

Secondly, what if f(x) is continuous at b and b is the end point of g(x) =
√f(x) such that g(x)'s domain is [b, +inf), we can say that g(x) is continuous from the right at x = b, correct? But not fully continuous (by fully continuous i mean the global limit exists; on the right and left).

Let's think about this. If the domain of function g is [b, inf), that means that g(x) is undefined for x<b, so that f(x)<0 for x < b. So you're assuming that f is continuous at b on its domain, right? The theorem would not apply, since f would not be continuous in an open interval around b; I think g will be continuous from the right, but I'd need to think more than I have to be sure I'm not missing anything, since they chose not to include this case in the theorem. But certainly g can't be continuous from the left, if the left is not in its domain, so that is unquestionable.
 
Thanks for clearing things out, yeah actually absolute value function is the best instance for that. And yup you said it, I was confusing what's true of f(x) > 0 with what's true of x
 
Top