Limits are the foundation of calculus. Before getting started with calculus, one must have the knowledge of limits. Limit of a function defines its nature, when the function approaches to a particular point. More specifically, limit of a function or a sequence is the value of that function or sequence, when it gets closer to a specified point. Limit of a function (or a sequence) is NOT its value at a definite point, rather than, it is the value of a function when it approaches nearest to a definite point.

Formal definition of limits is known as $\epsilon$ (epsilon), $\delta $ (delta) definition of the limits which states that:
Let $f$ be a function defined over an open interval which contains a value "$a$" in it.
Then, $f(x)$ equals $L$ when $x$ tends to "$a$" or we write
$\lim_{x \rightarrow a} f(x)$ = $L$
If for every real number $\epsilon > 0$, there exists a real number $\delta > 0$,
such that $\left  f(x)  L \right  < \epsilon $,
for all $0 < \left  x  a \right  < \delta$
Explanation: Let us assume a function f(x) which is shown in the following diagram:
Definition of limits says that the value to $f(x)$ is $L$, when $x$ is nearest to $a$. This means, when value of $x$ lies within the interval $(a  \delta, a + \delta)$ (where, $\delta$ is a real and positive number), eventually the value of $f(x)$ lies within the interval $(L  \epsilon, L + \epsilon)$ (where, $\epsilon$ is a real and positive number).
Limit of a function is known as its output value, when input approaches to a particular point. Let us consider a function $f(x)$ which is defined upon an open interval, then limit of $f(x)$ becomes $L$, when $x$ tends to "$a$" which is contained in given interval. We can write that:
$\lim_{x \rightarrow a}f(x)$ = $L$
Let us consider an easy example:
Solved Example
Question: Calculate $\lim_{x \rightarrow 2}(x^{2} + 1)$.
Solution:
Solution:
Substituting value of x = 2 in the given function,
$\lim_{x \rightarrow 2}(x^{2} + 1)$
$ = 2^{2} + 1 = 5$
$\lim_{x \rightarrow 2}(x^{2} + 1)$
$ = 2^{2} + 1 = 5$
Limit of a sequence defines convergence and divergence of a sequence. Let us consider a sequence ${x_{n}}$ over the set of real numbers, then ${x_{n}}$ converges to $L$, when n approaches to a sufficiently large number. In other words,
$\lim_{n \rightarrow \infty }{x_{n}}$ = $L$
If for every real number $\epsilon > 0$, there exists a natural number n $(n \geq N)$, such that $\left  x_{n}  L \right  < \epsilon$.
• If a sequence gives a finite value as it approaches to infinity, then the sequence is convergent or said to converge to that finite value.
• If a sequence gives an infinite value as it approaches to infinity, then the sequence is divergent or said to diverge.
Let us consider an easy example:
Solved Example
Question: Determine if the sequence ${\frac{1}{n}}$, where $n\ \epsilon\ N$, converges or diverges.
Solution:
Solution:
Sequence ${\frac{1}{n}}$ = 1, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, .....
$\lim_{n \rightarrow \infty } $$\frac{1}{n}$ = $\frac{1}{\infty}$ = $0$
The given sequence converges to zero.
$\lim_{n \rightarrow \infty } $$\frac{1}{n}$ = $\frac{1}{\infty}$ = $0$
The given sequence converges to zero.
Solved Examples
Question 1: Calculate $\lim_{x \rightarrow 1}$$\frac{x(4x^{2}  9)}{2x  3}$.
Solution:
Solution:
$\lim_{x \rightarrow 1}$$\frac{x(4x^{2}  9)}{2x  3}$
= $\lim_{x \rightarrow 1}$$\frac{x((2x)^{2}  3^{2})}{2x  3}$
= $\lim_{x \rightarrow 1}$$\frac{x((2x)^{2}  3^{2})}{2x  3}$
= $\lim_{x \rightarrow 1}$$\frac{x(2x + 3)(2x  3)}{2x  3}$
= $\lim_{x \rightarrow 1} \times (2x + 3)$
= $1 \times (2 \times 1 + 3)$
= 5
= $\lim_{x \rightarrow 1}$$\frac{x((2x)^{2}  3^{2})}{2x  3}$
= $\lim_{x \rightarrow 1}$$\frac{x((2x)^{2}  3^{2})}{2x  3}$
= $\lim_{x \rightarrow 1}$$\frac{x(2x + 3)(2x  3)}{2x  3}$
= $\lim_{x \rightarrow 1} \times (2x + 3)$
= $1 \times (2 \times 1 + 3)$
= 5
Question 2: A function f is defined as:
$f(x)=\left\{\begin{matrix}
x^{2} &if\ x\geqslant 3 \\
2x^{3}7 &if\ x< 3
\end{matrix}\right.$
Find the value of $\lim_{x\rightarrow 3}f(x)$ and $\lim_{x\rightarrow 2}f(x)$.
Solution:
$f(x)=\left\{\begin{matrix}
x^{2} &if\ x\geqslant 3 \\
2x^{3}7 &if\ x< 3
\end{matrix}\right.$
Find the value of $\lim_{x\rightarrow 3}f(x)$ and $\lim_{x\rightarrow 2}f(x)$.
Solution:
$\lim_{x\rightarrow 3}f(x)$ = $\lim_{x\rightarrow 3}x^{2}$
= $3^{2}$
= 9
$\lim_{x\rightarrow 2}f(x)$ = $\lim_{x\rightarrow 2}(2x^{3}  7)$
= $2(2)^{3}  7$
= 16  7
= 9
= $3^{2}$
= 9
$\lim_{x\rightarrow 2}f(x)$ = $\lim_{x\rightarrow 2}(2x^{3}  7)$
= $2(2)^{3}  7$
= 16  7
= 9
Question 3: Determine whether the following sequence is convergent:
$\frac{1}{3}$, $\frac{1}{5}$, $\frac{1}{7}$, $\frac{1}{9}$, .....
Solution:
$\frac{1}{3}$, $\frac{1}{5}$, $\frac{1}{7}$, $\frac{1}{9}$, .....
Solution:
Given sequence is
$\frac{1}{3}$, $\frac{1}{5}$, $\frac{1}{7}$, $\frac{1}{9}$, .....
n^{th} term of the sequence would be:
$\left \{ x_{n} \right \}$ = $\frac{1}{2n + 1}$
$\lim_{n \to \infty }\left \{ x_{n} \right \} = \lim_{n \to \infty }$$\frac{1}{2n + 1}$
= $\frac{1}{\infty}$ = 0
Therefore, the given sequence is convergent.
$\frac{1}{3}$, $\frac{1}{5}$, $\frac{1}{7}$, $\frac{1}{9}$, .....
n^{th} term of the sequence would be:
$\left \{ x_{n} \right \}$ = $\frac{1}{2n + 1}$
$\lim_{n \to \infty }\left \{ x_{n} \right \} = \lim_{n \to \infty }$$\frac{1}{2n + 1}$
= $\frac{1}{\infty}$ = 0
Therefore, the given sequence is convergent.