# Value of a function definition help

pivoxa15
Is the value of a function defined as the limit as x approaches each value in the domain of f(x)?

What happens if at a specific x, you have f(x)-> 0/0? is there usually a limit for which f(x) approaches in this situation?

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Homework Helper
Gold Member
Q1: no

Q2: a limit of the form 0/infinity is equal to 0.

pivoxa15
What should the answer to Q1 be?

WIth Q2. I am sorry, I stated it wrongly. It should be 0/0. I have corrected it in the OP.

Sourabh N
actually 0/0 is undefined. it can take different values in different situations.

example:
suppose f(x)= x^2/x
when u take limit for x->0 it becomes 0/0 .
but when u cancel x it gives ans as 0.

suppose f(x)= x/x^2
directly taking x-> gives again 0/0 .
but after cancellation it gives ans as infinity.

pivoxa15
So f(0)=1 when f(x)=x/x so 0/0 in this case equals to 1.

but 1/0 is undefined also isn't it? what is f(x)=1/x when x=0? There isn't one particular value for 1/0 like there was for 0/0 in the first situation is there? If so why?

Staff Emeritus
Gold Member
So f(0)=1 when f(x)=x/x
Yes
so 0/0 in this case equals to 1.
No! 0/0 is undefined. Always! It's not "equal to" any number.

but 1/0 is undefined also isn't it?
Yes.
what is f(x)=1/x when x=0?
It's not defined for x =0. You just said so yourself.
There isn't one particular value for 1/0 like there was for 0/0 in the first situation is there? If so why?
You're talking gibberish. 1/0 doesn't have a value. In any 'situation'.

pivoxa15
Yes No! 0/0 is undefined. Always! It's not "equal to" any number.

Yes. It's not defined for x =0. You just said so yourself. You're talking gibberish. 1/0 doesn't have a value. In any 'situation'.

But 0/0 can in some situations like x/x=1 where you take the limit as x->0. If you define any function with the limit as x approaches each specific point in the domain than you can define situations like x/x when x=0 couldn't you?

Homework Helper
Is the value of a function defined as the limit as x approaches each value in the domain of f(x)?

What happens if at a specific x, you have f(x)-> 0/0? is there usually a limit for which f(x) approaches in this situation?
Your original question doesn't make sense. If x is in the domain of a function f, then, by definition of "domain", the value of the function, f(x), is defined there. Do you mean the limit of the function? It is quite possible for a function value to be defined and the limit not exist there.
For example, f(x)= 0 if x is rational and 1 if x is irrational is defined for all real numbers but the limit does not exist anywhere.

So f(0)=1 when f(x)=x/x so 0/0 in this case equals to 1.
NO! If f(x)= x/x then its domain is "all real numbers except 0". f(0) is not defined. On the other hand, its limit as x goes to 0 IS defined and is 1. This is an example in which the limit is defined at a "limit point" of the domain that is not in the domain.

pivoxa15 said:
But 0/0 can in some situations like x/x=1 where you take the limit as x->0. If you define any function with the limit as x approaches each specific point in the domain than you can define situations like x/x when x=0 couldn't you?
NO again! You are confusing the limit with the value of the function. That's why your Calculus teacher is always careful to say things like
$$\frac{x^2- 4}{x-2}= x+ 2$$ as long as x is not equal to 2.

y= x+ 2 is defined for all x. Its graph is a straight line. (x2-4)/(x-2) is not defined at x= 2. Its graph is a straight line with a hole at (2, 4).

pivoxa15
NO again! You are confusing the limit with the value of the function. That's why your Calculus teacher is always careful to say things like
$$\frac{x^2- 4}{x-2}= x+ 2$$ as long as x is not equal to 2.

y= x+ 2 is defined for all x. Its graph is a straight line. (x2-4)/(x-2) is not defined at x= 2. Its graph is a straight line with a hole at (2, 4).

You have raised the issue that was the reason why I created this thread. Are you claiming f(x)=x/x does not equal to g(x)=1? Hence f(x) does not equal to g(x)? If so then it means g(x) is analytic everywhere in the domain of x but f(x) is not analytic everywhere? This would also mean that if you had a function g(x)=1 then it would be wrong to multiply the top and bottom by an arbitary x.

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Sourabh N
You have raised the issue that was the reason why I created this thread. Are you claiming f(x)=x/x does not equal to g(x)=1?

f(x)=x/x = 1 = g(x) is true only when x is not equal to zero. What I (& u) calculated for other functions was the value which they approach on taking the limit. But they don't take those values actually.

Hence f(x) does not equal to g(x)? If so then it means g(x) is analytic everywhere in the domain of x but f(x) is not analytic everywhere? This would also mean that if you had a function g(x)=1 then it would be wrong to multiply the top and bottom by an arbitary x.

As long as x is not equal to zero u can divide it with any expression.

ZioX
You have raised the issue that was the reason why I created this thread. Are you claiming f(x)=x/x does not equal to g(x)=1? Hence f(x) does not equal to g(x)? If so then it means g(x) is analytic everywhere in the domain of x but f(x) is not analytic everywhere? This would also mean that if you had a function g(x)=1 then it would be wrong to multiply the top and bottom by an arbitary x.

You're being very careless. Functions are not arbitrary maps, they have DOMAINS! f(x)=x/x has natural domain (the largest one possible) of R/{0} yet g(x) has natural domain R. If two functions do not share the same domain they cannot be equal.

f(0) is undefined until you actually define it as something. For example, we can have the function f(x)=x/x when x is not zero, and 1 when x=0 with domain R. This is the same as g(x)=1 with domain R.