I got confused on this

Hi
I'm really sorry that I didn't write anything about the way I solved it and why I've got confused in it.
In fact I'm Arabian and the problem is in Arabic so I tried hard to translate it to English (and that's why I didn't post the real photo) but I will try now to explain my idea.
I know that If x equal zero or f(x) equals to zero the square root of the limit will be non exist
But here what is given in the problem in that the limit of the function equal zero
So I thought that if the whole limit equal to zero then when we will find the square root of the limit it will be the square root of zero
Which is zero
My teacher didn't believe me and I have no prove for my idea if it was wrong or right!
Plus this problem was in the final exam and the answer is not existed!
 
If your teacher said that the answer is "d) does not exist" then they're wrong. The answer is "c) 0" no doubt about it. You were correct, but you can be more formal in your answer. Start by noting the properties of limits of composite functions:

\(\displaystyle \lim_{x \to a} g(f(x)) = g(\lim_{x \to a} f(x))\)

This identity requires two conditions to hold true. One, \( \lim\limits_{x \to a} f(x) \) must exist (i.e. it's equal to some real number \(L\)); Two, \( g(x) \) must be continuous at that point \( x = L \).

In your problem, you have \( g(x) = \sqrt{x} \). \( f(x) \) is left unspecified, but the text does give you exactly the first condition. Therefore, it must be the case that:

\(\displaystyle \lim_{x \to a} \sqrt{(f(x)} = \sqrt{\lim_{x \to a} f(x)} = \sqrt{0} = 0\)

The only reason why it might seem like this isn't the case is by considering a function such as the one pka suggested \( f(x) = x^3 \). In that case, it may seem like \(\displaystyle \lim_{x \to 0} \sqrt{x^3}\) doesn't exist because the function itself fails to exist for \(x < 0\). However, this is where we encounter a bit of a loophole, in that limits of functions with restricted domains need not always be two-sided. It's true that:

\(\displaystyle \lim_{x \to 0^{-}} \sqrt{x^3}\) doesn't exist

But the overall limit is, in this specific case, equal to the one-sided limit approaching from above. This occurs for the limit approaching any constant \(a\) for any f(x) that is undefined on the interval \(\displaystyle (a-\delta, \: a)\) for all \( \delta>0 \).

We know that \( f(x) = \sqrt{x^3} \) is undefined for negative inputs, so we can guarantee that \( f(x) \) is undefined on the interval \(\displaystyle (-\delta, \: a)\) for all \( \delta>0 \) and therefore:

\(\displaystyle \lim_{x \to 0} \sqrt{x^3} = \lim_{x \to 0^{+}} \sqrt{x^3}\)
 
My understanding is that calculus books often give a somewhat simplified definition of limit, requiring that the function be defined in an open interval containing a. By that definition, the limit seems not to exist. We need to see the book's definition -- and even then, we will probably object to it.

I think the teacher owes the class an explanation; also, it would be wrong to put this on a final if the issue was not discussed in class with a somewhat similar example. Just rejecting the student's argument without a positive counterargument is bad math.

On the other hand, what if [MATH]f(x) = -x^2[/MATH]?
 
ksdhart2 said:
If your teacher said that the answer is "d) does not exist" then they're wrong. The answer is "c) 0" no doubt about it. You were correct, but you can be more formal in your answer. Start by noting the properties of limits of composite functions:

\(\displaystyle \lim_{x \to a} g(f(x)) = g(\lim_{x \to a} f(x))\)

This identity requires two conditions to hold true. One, \( \lim\limits_{x \to a} f(x) \) must exist (i.e. it's equal to some real number \(L\)); Two, \( g(x) \) must be continuous at that point \( x = L \).

In your problem, you have \( g(x) = \sqrt{x} \). \( f(x) \) is left unspecified, but the text does give you exactly the first condition. Therefore, it must be the case that:

\(\displaystyle \lim_{x \to a} \sqrt{(f(x)} = \sqrt{\lim_{x \to a} f(x)} = \sqrt{0} = 0\)

The only reason why it might seem like this isn't the case is by considering a function such as the one pka suggested \( f(x) = x^3 \). In that case, it may seem like \(\displaystyle \lim_{x \to 0} \sqrt{x^3}\) doesn't exist because the function itself fails to exist for \(x < 0\). However, this is where we encounter a bit of a loophole, in that limits of functions with restricted domains need not always be two-sided. It's true that:

\(\displaystyle \lim_{x \to 0^{-}} \sqrt{x^3}\) doesn't exist

But the overall limit is, in this specific case, equal to the one-sided limit approaching from above. This occurs for the limit approaching any constant \(a\) for any f(x) that is undefined on the interval \(\displaystyle (a-\delta, \: a)\) for all \( \delta>0 \).

We know that \( f(x) = \sqrt{x^3} \) is undefined for negative inputs, so we can guarantee that \( f(x) \) is undefined on the interval \(\displaystyle (-\delta, \: a)\) for all \( \delta>0 \) and therefore:

\(\displaystyle \lim_{x \to 0} \sqrt{x^3} = \lim_{x \to 0^{+}} \sqrt{x^3}\)
Click to expand...
Thanks a lotttttt
 
If f(x) was approaching 0 through ONLY positive numbers than under any definition I would think the answer to your question will be 0. How could it be non exist???? (my question is meant for your teacher)
 
Jomo said:
If f(x) was approaching 0 through ONLY positive numbers than under any definition I would think the answer to your question will be 0. How could it be non exist???? (my question is meant for your teacher)
Click to expand...
I don't know why she think in that way but I guess because of the answer form said so!
 
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