# 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 $0 < \left | x - a \right |< \delta$

## 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:

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

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

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 $y = f(x)$.

## 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 $\varepsilon>0$

$\left | \frac{1}{x^{2}-0} \right |<\varepsilon$

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.

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.