Existence of a limit: concept

When it comes to limit we say $\lim_{x \rightarrow a} f(x)$ exists if and only if

$$\tag{1} \lim_{x \to a+} f(x) = \lim_{x \to a-} f(x)$$ (equating both sides of limit)

It simply implies that $f(x)$ should approach to same value from both sides. I derive the limit values using the basic limit theorems as in this link. So far fine. When it comes to one sided limits separately, I apply the same theorems with left or right direction in mind. If I am able to derive a reasonable value from theorem I can say that one sided limit exists for that particular direction. But I don't think,it is sound in conceptual thinking. I am looking for a mathematical definition of existence of One sided limit as in equation (1) for general limit

What is the mathematical explanation for the existence of one sided limits such as $$\tag{2} \lim_{x \to a+} f(x)$$ , $$\tag{3} \lim_{x \to a-} f(x)$$

Is it possible to have a example where

$$\tag{4} \lim_{x \to -a} f(x) , \lim_{x \to a} f(x)$$ do not exists when we consider it separately

Further explanation with example
A function f(x) is said to be continuous on the right of a if it satisfies conditions

1) $\lim_{x \to +a} f(x)$ exits
2) $f(a)$ is defined
3) $\lim_{x \to +a} f(x)$ =f(a)

DOUBT
How do we decide $\lim_{x \to +a} f(x)$ exits or not?. We know already how to check existence with $\lim_{x \to a} f(x)$ as in equation(1)

• Look at the folowing function $f$, such that. $f(x)=x+1$ for $x>=0$ and $f(x)=x$ for $x<0$. Try to find one-sided limits at 0. Using the definition of one-sided limit you can easily prove that limit from the right is 1 and limit from the left is 0. – Юрій Ярош May 5 '18 at 6:01
• Thanks for the suggestion . I have edited more explanation – Nirvana May 5 '18 at 10:36

Well, there are many theorems which can be used to evaluate one sided limits (in exactly the same manner as they are used for the usual two sided limits). A successful evaluation of the limit means that it exists. Apart from this there are few theorems which allow us to infer the existence of a limit without performing any evaluation.

One such theorem is the following :

Theorem: Let $f$ be increasing on some interval of type $(a-h, a)$ and let $f$ be also bounded above on this interval. Then $\lim_{x\to a^{-}} f(x)$ exists. A similar result holds for functions decreasing and bounded below on $(b, b+h)$ and the limit $\lim_{x\to b^{+}} f(x)$.

Consider the Heaviside function. Note that as we approach the point $c$ from the right, the limit is $1$, whereas from the left, it is $0$. Hope this is an illuminating example. The limit does not exist if we approach from both sides.

• Thanks for the suggestion . I have edited more explanation – Nirvana May 5 '18 at 10:36

Firstly, One sided limit simply means that the domain of the function is restricted to one of the sides of a where x is approaching in a function f(x).

Now your question on how to know that the one sided limit exists or not then see if f(x)=1/x is your function
Now let x approach 0 from right hand side. So, x=1/10 : f(x) = 10

  x= 1/10000   :f(x) =10000


So as you are going near to 0 the value of the function is going bigger and bigger even beyond any limit. So off course limit doesn't exist but you must know what it means by non existence of a limit. :)

I hope I understand your question correctly. So you are searching for the definition of one-sided limit. So $f(x)$ approaches the limit $L$ from the positive(right) side as $x$ aproaches $a$ if and only if for any $\epsilon>0$ there exists $\delta>0$ such that from the inequality $0<x-a<\delta$ we can conclude that $|f(x)-L|<\epsilon$. Similarly, $f(x)$ approaches the limit $L$ from the negative(left) side as $x$ aproaches $a$ if and only if for any $\epsilon>0$ there exists $\delta>0$ such that from the inequality &0