The concept of limit plays a vital role in the study of calculus. The theory of limits which forms the basic for the deep knowledge of continuity, which will lay the foundation for the study of differentiability and differentiation of the function. In this section we will learn more about limits.

Let $f(x)$ be a function defined for all $x$ in the neighborhood of '$a$' except possibly at '$a$'. then '$l$' is said to be the limiting value of $f(x)$ as $x$ tends to $a$. If the numerical difference between $f(x)$ and $l$ can be maid as small as we please by taking $x$ sufficiently close to 'a' but not equal to 'a'.
we write this as,

$\lim_{x \to a}f(x)$ = $l$

Left hand limit of a function: Let f(x) be a function of x. If $f(x)$ approaches to $l$ as $x \to a^{-}$, then $l$ is called as the left hand limit of the function and we write it as,

$\lim_{x \to a^{-}}f(x)$ = $l$

Right hand limit of a function: Let f(x) be a function of x. If $f(x)$ approaches to $l$ as $x \to a^{+}$, then $l$ is called as the left hand limit of the function and we write it as,

$\lim_{x \to a^{+}}f(x)$ = $l$

$\lim_{x \to a}f(x)$ exists if and only if $\lim_{x \to a^{-}}f(x)$ and $\lim_{x \to a^{+}}f(x)$ both exist and are equal.

Basic formulas :
$\lim_{x \to a}c$ = $c$ , where 'c' is a constant.

$\lim_{x \to 0^{+}}$$\frac{1}{x}$ = $\infty$

$\lim_{x \to 0^{-}}$$\frac{1}{x}$ = $-\infty$

$\lim_{x \to \infty}$$\frac{1}{x}$ = $0$

$\lim_{x \to -\infty}$$\frac{1}{x}$ = $0$

$\lim_{x \to 0}$$\frac{sin x}{x}$ = $1$

$\lim_{x \to 0}$$\frac{cos x-1}{x}$ = $0$

$\lim_{x \to \infty}(1+$$\frac{1}{x}$$)^{x}$ = $e$

Solved Examples

Question 1: Evaluate $\lim_{x \to -3}$ $\frac{x^{2}-x-6}{x^{2}+x-2}$?
Solution:
 
$\lim_{x \to -3}$ $\frac{x^{2}-x-6}{x^{2}+x-2}$ = $\lim_{x \to -3}$ $\frac{(x+2)(x-3)}{(x+2)(x-1)}$

= $\lim_{x \to -3}$ $\frac{(x-3)}{(x-1)}$

= $\frac{(-3-3)}{(-3-1)}$

= $\frac{-6}{-4}$

= $\frac{3}{2}$
 

Question 2: Evaluate $\lim_{x \to 2}$ $\frac{x^{2}-9}{x-3}$?
Solution:
 
$\lim_{x \to 2}$ $\frac{x^{2}-9}{x-3}$ = $\lim_{x \to 2}$ $\frac{(x-3)(x+3)}{(x-3)}$

= $\lim_{x \to 2}$ $\frac{(x+3)}{(1)}$

= $\frac{(2+3)}{(1)}$

= $\frac{5}{1}$

= $5$
 

Question 3: Show that $lim_{x \to 0}$$\frac{x}{tan x}$= $1$.
Solution:
 
LHS = $lim_{x \to 0}$$\frac{x}{tan x}$

= $lim_{x \to 0}$$\frac{x}{\frac{sin x}{cos x}}$

= $lim_{x \to 0}$$\frac{x \cos x}{sin x}$

= $lim_{x \to 0}$$\frac{x}{sin x}$. $lim_{x \to 0}$$\frac{cos x}{1}$

From the fact its clear that $lim_{x \to 0}$$\frac{x}{sin x}$ =$1$

= $lim_{x \to 0}$$\frac{x}{sin x}$. $lim_{x \to 0}$$\frac{cos x}{1}$ = $1.1$

= $1$ = RHS
 

Let $l$ be a real number. The limit of sequence {$a_{n}$} is $l$. written as $\lim_{x \to \infty}f(x)$ = $l$
if for each $\varepsilon > 0$, there exist M > 0 such that $|a_{n} – l| < \varepsilon$ where n > M.

If the limit $l$ of the sequence exists, then the sequence converges to $l$. If the limit of the sequence dose not exist, then the sequence diverges.