# Limits in Maths Notes

## Definition of Limits

In mathematical terms, the limit of a function $f(x)$ as $x$ approaches a point $c$ is given by
$\lim_{{x \to c}} f(x) = L$

Here, $L$ is the value that $f(x)$ approaches as $x$ gets closer to $c$.
The limit of f (x) as x tends to x is to be thought of as the value f (x) should assume at x = c

Another Definition
The formal definition of a limit is as follows: Let $f(x)$ be a function defined on some open interval containing $c$, except possibly at $c$ itself. We say that the limit of $f(x)$ as $x$ approaches $c$ is $L$, written as
$\lim_{{x \to c}} f(x) = L$

if for every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $0 < |x - c| < \delta$, then $|f(x) - L| < \epsilon$.

## Left-hand Limit

The left-hand limit of $f(x)$ as $x$ approaches $c$ is denoted by
$\lim_{{x \to c^-}} f(x)$

It is the expected value of f at x = a given the values of f near x to the left of a.

## Right-hand Limit

The right-hand limit of $f(x)$ as $x$ approaches $c$ is denoted by
$\lim_{{x \to c^+}} f(x)$

## Existence of Limit

For a limit to exist, the left-hand and right-hand limits must be equal, i.e.,
$\lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x) = \lim_{{x \to c}} f(x)$

## Algebra of Limits

1. Limit of a Sum: Limit of sum of two functions is sum of the limits of the functions
$\lim_{{x \to c}} [ f(x) + g(x) ] = \lim_{{x \to c}} f(x) + \lim_{{x \to c}} g(x)$
2.Limit of Difference: Limit of difference of two functions is sum of the limits of the functions
$\lim_{{x \to c}} [ f(x) - g(x) ] = \lim_{{x \to c}} f(x) - \lim_{{x \to c}} g(x)$
3. Limit of a Product : Limit of product of two functions is product of the limits of the functions
$\lim_{{x \to c}} [ f(x) \times g(x) ] = \lim_{{x \to c}} f(x) \times \lim_{{x \to c}} g(x)$
4 if g(x)=k,then
$\lim_{{x \to c}} [ f(x) \times k ] = k \lim_{{x \to c}} f(x)$
5. Limit of a Quotient: Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero)
$\lim_{{x \to c}} \frac{f(x)}{g(x)} = \frac{\lim_{{x \to c}} f(x)}{\lim_{{x \to c}} g(x)}$ (provided $\lim_{{x \to c}} g(x) \neq 0$)

## Types of Limits

Finite Limits These are limits where the function approaches a particular real number as $x$ approaches a specific value. For example,
$\lim_{{x \to 2}} (x^2 - 4) = 0$

## Infinite Limits

Here, the function approaches infinity as $x$ approaches a specific value. For example,
$\lim_{{x \to 0}} \frac{1}{x^2} = \infty$

## Limits at Infinity

These limits describe the behavior of a function as $x$ goes to infinity. For example,
$\lim_{{x \to \infty}} \frac{1}{x} = 0$

## Limit of Polynomial Function

We know that
$\lim_{{x \to a}} x = a$
Now
$\lim_{{x \to a}} x^2 = \lim_{{x \to a}} x.x = \lim_{{x \to a}} x \lim_{{x \to a}} x= a.a=a^2$
Similarly
$\lim_{{x \to a}} x^n = a^n$

A polynomial function f(x) in one variable x is an algebraic expression in x term as
$f(x)=a_n x^n+a_(n-1) x^(n-1)+a_(n-2) x^(n-2)+⋯………+ax+a_0$
Where $a_n,a_{n-1},....,a,a_0$ are constant and real numbers and $a_n$ is not equal to zero
Limit is calculated as
$\lim_{{x \to a}} f(x) =\lim_{{x \to a}} a_n x^n+a_(n-1) x^(n-1)+a_(n-2) x^(n-2)+⋯………+ax+a_0$
$=a_n \lim_{{x \to a}} x^n + a_(n-1) \lim_{{x \to a}} x^(n-1)+ a_(n-2) \lim_{{x \to a}} x^(n-2)+⋯………+ a \lim_{{x \to a}} x+a_0 \lim_{{x \to a}}$
$=a_n a^n + a_(n-1) a^(n-1)+ a_(n-2) a^(n-2)+⋯………+ a_1 a+a_0 =f(a)$
Hence
$\lim_{{x \to a}} f(x) = f(a)$

## Limit of Rational Function

Rational Function is defined as
$f(x)=\frac {g(x)}{h(x)}$
Where g(x) and h(x) are polynomial function and $h(x) \ne 0$
$\lim_{{x \to a}} \frac {g(x)}{h(x)} = \frac { \lim_{{x \to a}} g(x)}{\lim_{{x \to a}} h(x)}= \frac {g(a)}{h(a)}$

Case A if h(a) =0 and g(a) =0
Now we can write
$g(x) = (x-a)^k g' (x)$
$h(x) = (x-a)^l h' (x)$
if k > l

$\lim_{{x \to a}} \frac {g(x)}{h(x)} = \frac { \lim_{{x \to a}} g(x)}{\lim_{{x \to a}} h(x)}=\frac { \lim_{{x \to a}}(x-a)^k g'(x)}{\lim_{{x \to a}} (x-a)^l h'(x)}= \frac {0.g(a)}{h(a)}=0$

If k< l
Limit is undefined
Case B - h(a) =0 and $g(a) \ne 0$
Then Limit is undefined

A general rule that needs to be kept in mind while evaluating limits is the following.
(i) First we check the value of f (a) and g(a).
(ii)If both are 0, then we see if we can get the factor which is causing the terms to vanish, i.e., see if we can write $f(x) = f_1(x) f_2(x)$ so that f1(a) = 0 and $f_2(a) \ne 0$.
(iii)Similarly, we write $g(x) = g_1(x) g_2(x)$, where $g_1(a) = 0$ and $g_2(a) \ne 0$.
(iv) Cancel out the common factors from f(x) and g(x) (if possible) and write
$\frac {f(x)}{g(x)} = \frac {p(x)}{q(x)}$

## Special Theorem

Theorem I
$\lim_{{x \to a}} \frac {x^n -a^n}{x-a} = na^{n-1}$

This is valid for when n is positive integers and n is any rational number and a is positive

Proof
$\frac {x^n -a^n}{x-a}= (x^{n–1} + x^{n–2} a + x^{n–3} a^2+ ... + x a^{n–2} + a^{n–1}$
Hence
$\lim_{{x \to a}} \frac {x^n -a^n}{x-a} = a^{n – l} + aa^{n–2} +. . . + a^{n–2} (a) +a^{n–l}= na^{n-1}$

Theorem II Let f and g be two real valued functions with the same domain such that $f (x) \leq g( x)$ for all x in the domain of definition, For some a, if both
$\lim_{{x \to a}} f(x)$ and $\lim_{{x \to a}} g(x)$
exists, then
$\lim_{{x \to a}} f(x) \leq \lim_{{x \to a}} g(x)$

Sandwitch Theorem
Let f, g and h be real functions such that
$f (x) \leq g( x) \leq h(x)$ for all x in the common domain of definition.
For some real number a, if
$\lim_{{x \to a}} f(x) =l=\lim_{{x \to a}} h(x)$
then
$\lim_{{x \to a}} g(x) =l$

## Limits of Trigonometric Function

we can easily the below relation geometrically
$cos x < \frac {sin x}{x} < 1$ for $0 < |x| < \frac {\pi}{2}$

Lets define the limit of two important Theorem based on Above Formula and Sandwitch theorem Theorem I
$\lim_{{x \to 0}} \frac {sin x}{x} =1$
Proof
$\lim_{{x \to 0}} cos x =1$
$\lim_{{x \to 0}} 1 =1$
Therefore from Sandwitch theorem
$\lim_{{x \to 0}} \frac {sin x}{x} =1$

Theorem II
$\lim_{{x \to 0}} \frac {1-cosx}{x}=1$
Proof
$\lim_{{x \to 0}} \frac {1-cosx}{x} =\lim_{{x \to 0}} \frac {2sin^2(x/2)}{x} =\lim_{{x \to 0}} \frac {sin^2(x/2)}{x/2}$
$=\lim_{{x \to 0}} \frac {sin(x/2)}{x/2} sin (x/2) = 1.0 =1$
Therefore
$\lim_{{x \to 0}} \frac {1-cosx}{x}=1$

Theorem III
$\lim_{{x \to 0}} \frac {tan x}{x}=1$
Proof
$\lim_{{x \to 0}} \frac {tanx}{x}=\lim_{{x \to 0}} \frac {sinx}{x cosx}= \lim_{{x \to 0}} \frac {sinx }{x} \lim_{{x \to 0}} \frac {1}{cos x}=1$
Therefore
$\lim_{{x \to 0}} \frac {tan x}{x}=1$

## How do I calculate a limit

(a)A limit is undefined if the function doesn't approach a specific value or if it approaches different values from the left and right.
(b) Limits can be calculated using a variety of methods including direct substitution, factorization, rationalization,Theorems stated abive and using L'Hôpital's Rule for indeterminate forms.

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