Formal and epsilon delta definition of Limit of a function with examples

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

Formal and epsilon delta definition of Limit of a function with examples
Limit notation of a function

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:

  1. \lim ” tells us that we are looking at the limit value of a function, not the function value of a function,
  2. 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”,
  3. f \left ( x \right ) ” tells us which function we are working with,
  4. 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

limit x tends to 2x equal to 10

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 > \varepsilon

Hence 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

Limit of a function graphically
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

limit x tends to 2x equal to 10

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 |<\varepsilon

i.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

  1. 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

  2. Is the limit of a function unique?

    Yes, the limit of a function is unique.
    Click here to see the proof of limit uniqueness.

  3. 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

  4. 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.

Read more: How to find the zeros of a quadratic function?

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