condition of existence of limits

Millesa

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Prove that the function f (x) = ∣x∣/x has no lmiit
when x → 0

ok.Does not exist limit when your value is zero or only when is ∞?

Does too need to check the continuity?
 
Prove that the function f (x) = ∣x∣/x has no lmiit
when x → 0 ok.Does not exist limit when your value is zero or only when is ∞?
Does too need to check the continuity?
Is it clear to that: \(\displaystyle \large{\mathop {\lim }\limits_{x \to {0^ + }} \frac{{|x|}}{x} = 1\quad \& \quad \mathop {\lim }\limits_{x \to {0^ - }} \frac{{|x|}}{x} = - 1\;?}\)
What does that tell you?
 
Prove that the function f (x) = ∣x∣/x has no lmiit
when x → 0

ok.Does not exist limit when your value is zero or only when is ∞?

Does too need to check the continuity?
Two comments:
1) I am not sure why you are talking about when x=∞
2) A function f(x) does NOT have to be defined at x=a for the lim as x goes to a to exist. You really need to see that!

Now answer pka question to get your answer. Of course if you do not understand why those limits were 1 and -1 then please ask why.
 
The limit "as x->0" depends only on the values of the function very close to 0. There is no reason to talk about any other values of x and certainly not 0! Also, a function does not have to be "continuous" or even have a value at x= a in order to have a limit there. In order that f(x) be continuous at x= a,
1) the limit of f(x) at x= a must exist.
2) the value of the function, f(a), must exist.
3) those must be equal.

So being continuous at a point is a much stronger condition than just having a limit there!

What you need here is the fact that "the limit, as x goes to a, of f(x)" exists if and only if the two limits "from above" and "from below" exist and are the same.

Since the limit here is at 0, "from above" means x> 0. If x is positive what can youi say about |x|/x? What is its limit? If x is negative, what can you say about |x|/x? What is its limit?
 
You need to understand the definition of |x| ! Do you?

|x| = x if x>0 and |x| = -x if x<0

So sometimes |x|/x =- x/x =1 and other times |x|/x = -x/x = -1

Now put ALL the pieces together. This is Calculus!
 
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