In Calculus, the limit of a function is a fundamental concept. With the help of the concept of the limit of a function, we can understand the behavior of a function f(x) near a point x.
In the previous chapter, we have learned about Function and now in this chapter, we will discuss the concept limit of a function.
During this discussion, our main focus will be on the following topics:
- the formal definition of limit of a function,
- epsilon delta definition of limit of a function with examples
But before that, we will understand the meaning of x tends to a ( x \to a ).
Meaning of x tends to a
To understand the meaning of x tends to a (i.e., x \to a ) we take the example of x \to 2 .
The symbol x \to 2 means that the variable x gradually approaches towards 2 either from the left of the right side of the point x=2 by assuming successive values according to some chosen law in such a manner that the numerical difference between the assumed value of x and 2 (i.e., the values of \left | x-2 \right | ) can be made less than any pre-assigned positive number \delta (an arbitrarily chosen number, however small).
For example, let the variable x assumes the successive values of the infinite sequence 1.9, 1.99, 1.999, ….. from the left of x=2 and let the pre-assigned positive number \delta = 0.001 (which may be chosen as small as we like).
Then taking x = 1.9999 we can make \left | x-2 \right | < \delta .
When x=1.9999
\left | x - 2 \right | = \left | 1.9999 - 2 \right | = 0.0001 < 0.001 < \delta .
Therefore \left | x - 2 \right | < \delta .
Again let the successive values of the infinite sequence 2.1, 2.01, 2.001, ….. from the right of x = 2 and let the pre-assigned positive number \delta = 0.001 (which may be chosen as small as we like).
Then when x = 2.0001
\left | x - 2 \right | = \left | 2.0001 - 2 \right | = 0.0001 < 0.001 < \delta .
Therefore \left | x - 2 \right | < \delta .
Definition of limit of a function
Let f be a real function defined on a domain D \subset \mathbb{R} .
In order that f may have a limit l \left ( l \epsilon \mathbb{R} \right ) at a point a , for x sufficiently close to a , f(x) should be arbitrarily close to l .
Formal definition of limit of a function
Let D \subset \mathbb{R} and f : D \rightarrow \mathbb{R} be a function and a be a limit point of D .
A real number l is said to be the limit of f at a if corresponding to any neighbourhood V of l there exists a neighbourhood W of a such that f(x) \epsilon V for all x \epsilon [ W - { a } ] \cap D
This is expressed by the symbol
\lim _{x \rightarrow a} f \left ( x \right ) = l
or,
f \left ( x \right ) \rightarrow l as x \rightarrow a
Epsilon delta definition of limit of a function
Let D \subset \mathbb{R} and f : D \rightarrow \mathbb{R} be a function and a be a limit point of D .
A real number l is said to be a limit of f at a if corresponding to a pre-assigned +ve \varepsilon there exists a positive \delta such that
l - \varepsilon < f \left ( x \right ) < l + \varepsilon
for all x \epsilon {N}'\left ( a, \delta \right )\cap D ,
where {N}'\left ( a, \delta \right ) = { x \epsilon \mathbb{R} : 0 < \left | x - a \right | < \delta } = \left ( a - \delta , a + \delta \right ) - { a }.
In simple words
For every positive number \varepsilon , no matter how small, there exists a positive \delta such that
\left | f\left ( x \right ) - l \right |< \varepsilon , whenever 0 < \left | x - a \right |< \delta
Read also: What is a function in Math? – Definition, Example, and graph
Limit notation
The limit notation in calculus is
So what does this limit notation mean?
What we can know from the limit notation
The limit notation has 4 parts and each of these parts has its own meaning:
- “ \lim ” tells us that we are looking at the limit value of a function, not the function value of a function,
- “ x \rightarrow a ” tells us what is the variable of the function and what it is approaching and we read it as “x tends to a”,
- “ f \left ( x \right ) ” tells us which function we are working with,
- “ l ” tells us the limit of the function f \left ( x \right ) i.e., what value the function is approaching.
For example, look at the limit
Here
- “ x \rightarrow 2 ” tells us x is approaching to 2 (i.e., x tends to 2 ),
- “ 5x ” is the function of which limit we have to find,
- “ 10 ” tells us 5x is approaching to 10 as x is approaching 2 i.e., 5x \rightarrow 10 as x \rightarrow 2 .
What is the negation of the statement \lim _{x \rightarrow a} f \left ( x \right ) = l
If f(x) does not tend to l as x \rightarrow a , then it is not true for every \varepsilon > 0 , there exist some \delta > 0 such that
\left | f\left ( x \right ) - l \right | < \varepsilon , \forall x \epsilon 0< \left | x - a \right | < \delta
\therefore if f(x) does not tend to l as x \rightarrow a , then there exist some \varepsilon > 0 such that for every \delta > 0 there is some x satisfying 0 < \left | x - a \right | < \delta but not satisfying \left | f\left ( x \right ) -l \right | < \varepsilon .
In short, if f(x) does not tend to l as x tends to a , then we shall be able to find an \varepsilon > 0 such that for every positive number \delta there exist some x for which
0 < \left | x - a \right | < \delta but 0 < \left | f \left ( x \right ) - l \right | \geq \varepsilon .
Example: Prove the statement using the epsilon delta definition of limit of a function that the function f, defined by f\left ( x \right ) = \sin \left ( \frac{1}{x} \right ) , when x \neq 0 and f(0) = 0 , does not approach 0 as x \rightarrow 0 .
Solution: We use epsilon delta definition of limit of a function to prove the statement.
Take \varepsilon = \frac{1}{2} .
If \delta be any positive number, then it is possible to find a positive integer n such that
\delta > \frac{2}{\left ( 4n + 1 \right ) \pi }or, \left ( 4n + 1 \right ) \pi > \frac{1}{\delta }
or, \left ( 2n + \frac{1}{2} \right ) \pi > \frac{1}{\delta }
or, 2n \pi + \frac{\pi}{2} > \frac{1}{\delta }
Then clearly, 0 < \left | x - 0 \right | < \delta , when x = \frac{1}{2n\pi + \frac{\pi }{2}} and
\left | f\left ( x \right ) - 0 \right | = \left | \sin \frac{1}{x} \right | = \left | \sin \left ( 2n\pi + \frac{\pi }{2} \right ) \right | = 1 > \varepsilonHence f(x) does not tend to zero as x \rightarrow 0 ;
i.e., it is not true that
\lim _{x\rightarrow 0} \sin \frac{1}{x} = 0 .
Epsilon delta definition of Limit of a function Graphically
The existence of limit l of f(x) as x \rightarrow a is shown on the graph
l and \varepsilon , being known, we can draw the horizontal lines y = l , y = l - \varepsilon and y = l + \varepsilon .
y \rightarrow l as x \rightarrow a , if only a \delta – neighbourhood of the point x = a can be found such that all points on the graph y = f(x) corresponding to the points x of the neighbourhood (excepting perhaps for the point x = a ) lie within the strip bounded by the lines y = l \pm \varepsilon and x = a \pm \delta .
The point P(a, l) may or may not belong to the graph of y = f(x) .
Read also: 48 Different Types of Functions and their Graphs [Complete list]
Difference between limit x tends to a f(x) and f(a)
The concept of limit of a function will be incomplete if we do not the difference between \lim _{x \to a} f \left ( x \right ) and f \left ( a \right ) .
The symbol \lim _{x \to a} f \left ( x \right ) stands for the value of f(x) when x is sufficiently close to a .
Here we want to know what happens to the value f(x) when x approaches a either from the left or from the right of the point x = a .
But we never consider what happens to f(x) when x is exactly equal to a .
On the contrary f(a) stands for the value of f(x) when x is exactly equal to a.
The value of f(a) is obtained using the definition of f(x) at x = a or else by the substitution a for x in the analytical expression for f(x) when it exists.
Is the limit of a function unique?
Theory: Prove that the limit \lim _{x \rightarrow a}f\left ( x \right ) , when exist, is unique.
Proof: If possible, let \lim _{x \rightarrow a}f\left ( x \right ) has two distinct limits l_{1} and l_{2} where l_{1} \neq l_{2} .
Let \varepsilon > 0 be any given positive number.
We can choose \delta > 0 suitable such that
\left | f(x) - l_{1}) \right | < \frac{\varepsilon }{2}, \forall x\epsilon 0 < \left | x - a \right | < \delta\left | f(x) - l_{2}) \right | < \frac{\varepsilon }{2}, \forall x\epsilon 0 < \left | x - a \right | < \delta .
Now
\left | l_{1} - l_{2} \right |
= \left | l_{1} - f(x) + f(x) - l_{2} \right |
\leq \left | l_{1} - f(x)) \right | + \left | f(x) - l^{} \right |
< \frac{\varepsilon }{2} + \frac{\varepsilon }{2}
= \varepsilon, \forall x \epsilon 0 < \left | x - a \right | < \delta
Thus \left | l_{1} - l_{2} \right | is less than any positive number (no matter how small we take). This is only possible if \left | l_{1} - l_{2} \right | = 0 i.e., if l_{1} = l_{2} .
Hence it is proved that limit \lim _{x \rightarrow a}f\left ( x \right ) , when exist, is unique.
Limit of a function examples with answers
Example 1: Prove the statement using the epsilon delta definition of limit of a function that \lim_{x \rightarrow 2} 5x = 10
Solution: Using the epsilon delta definition of limit of a function, we have
For any given \varepsilon > 0 , there exists \delta > 0 such that
\left | 5x - 10 \right | < \varepsilon , whenever 0 < \left | x - 2 \right | < \delta .
See that \left | 5x - 10 \right | < \varepsilon \Rightarrow \left | x - 2 \right | < \frac{\varepsilon }{5} .
Therefore choosing \delta = \frac{\varepsilon }{5} , epsilon delta definition of limit of a function is satisfied.
Therefore
Read more: How to find the zeros of a function?
Example 2: Prove the statement using the epsilon delta definition of limit of a function that \lim _{x \rightarrow 3}\frac{1}{x} = \frac{1}{3} .
Solution: Using the epsilon delta definition of limit of a function, we get
For any given \varepsilon > 0 , there exists \delta > 0 such that
\left | \frac{1}{x} - \frac{1}{3} \right | < \varepsilon , whenever 0 < \left | x - 3 \right | < \delta
See that \left | \frac{1}{x} - \frac{1}{3} \right | < \varepsilon
i.e., \frac{1}{3} - \varepsilon < \frac{1}{x} < \frac{1}{3} + \varepsilon
i.e., \frac{1 - 3 \varepsilon }{3} < \frac{1}{x} < \frac{1 + 3 \varepsilon }{3}
i.e., \frac{3}{1 + 3 \varepsilon } < x < \frac{3}{1 - 3 \varepsilon }
i.e., \frac{3}{1 + 3 \varepsilon } -3 < x -3 < \frac{3}{1 - 3 \varepsilon } -3
i.e., \frac{-9 \varepsilon }{1 + 3 \varepsilon } < x -3 < \frac{9 \varepsilon }{1 - 3 \varepsilon }
Take \delta _{1} = \frac{9 \varepsilon }{1 + 3 \varepsilon }, \delta _{2} = \frac{9 \varepsilon }{1 - 3 \varepsilon } .
Let \delta = min { \delta_{1}, \delta_{2} } = \delta_{1} = \frac{9 \varepsilon }{1 + 3 \varepsilon } .
So \lim_{x \rightarrow 3} \frac{1}{x} = \frac{1}{3} .
Here the epsilon delta definition of limit of a function is satisfied.
This completes the proof.
Example 3: Does the limit exist \lim_{x \rightarrow \infty } \frac{1}{x^{2}} = 0 ? Explain your answer.
Solution: Using epsilon delta definition of limit of a function, we get
For, given any \varepsilon>0
\left | \frac{1}{x^{2}-0} \right |<\varepsiloni.e., \left | \frac{1}{x^{2}} \right |<\varepsilon
i.e., \frac{1}{x^{2}} < \varepsilon
i.e., x^{2} > \frac{1}{\varepsilon}
i.e., x > \sqrt{\frac{1}{\varepsilon}}
i.e., x>\frac{1}{\sqrt{\varepsilon}}
which shows that the epsilon delta definition of limit of a function is satisfied.
Thereforethe limit \lim_{x \rightarrow \infty } \frac{1}{x^{2}}=0 exists.
Frequently asked questions on Epsilon Delta definition of Limit
-
What do you understand by limit of a function?
A real number l is said to be the limit of a function f(x) at x = a if corresponding to a pre-assigned +ve \varepsilon there exists a positive \delta such that
\left | f\left ( x \right ) - l \right |< \varepsilon , whenever 0 < \left | x - a \right |< \delta -
Is the limit of a function unique?
Yes, the limit of a function is unique.
Click here to see the proof of limit uniqueness. -
Can the limit of a function be zero?
Yes, the limit of a function be zero.
For example, let f(x) = x - 1 .
We can see that \lim _{x \to 1} \left ( x - 1 \right ) = 0 -
Can limit of a function be infinity?
Yes, the limit of a function be infinity.
For example, let f(x) = \frac{1}{x^2} be a function.
Then \lim _{x \to 0 } \frac{1}{x^2} = \infty
which shows that the limit of a function be infinity.
We hope you understand the definition of limit of a function, formal definition of limit, epsilon delta definition of limit.
Furthermore, if you have any questions related to the topic epsilon delta definition of limit of a function, feel free to ask in the comment section. We will definitely answer you.