\[ \lim_{{x \to c}} f(x) = L \]

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

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 \( |f(x) - L| < \epsilon \).

\[ \lim_{{x \to c^-}} f(x) \]

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

\[ \lim_{{x \to c^+}} f(x) \]

\[ \lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x) = \lim_{{x \to c}} f(x) \]

\( \lim_{{x \to c}} [ f(x) + g(x) ] = \lim_{{x \to c}} f(x) + \lim_{{x \to c}} g(x) \)

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 \( \lim_{{x \to c}} g(x) \neq 0 \))

\[ \lim_{{x \to 2}} (x^2 - 4) = 0 \]

\[ \lim_{{x \to 0}} \frac{1}{x^2} = \infty \]

\[ \lim_{{x \to \infty}} \frac{1}{x} = 0 \]

\[ \lim_{{x \to a}} x = a \]

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

\[ \lim_{{x \to a}} f(x) = f(a) \]

$f(x)=\frac {g(x)}{h(x)}$

Where g(x) and h(x) are polynomial function and $h(x) \ne 0$

\[ \lim_{{x \to a}} \frac {g(x)}{h(x)} = \frac { \lim_{{x \to a}} g(x)}{\lim_{{x \to a}} h(x)}= \frac {g(a)}{h(a)} \]

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

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

$\frac {f(x)}{g(x)} = \frac {p(x)}{q(x)}$

\[ \lim_{{x \to a}} \frac {x^n -a^n}{x-a} = na^{n-1} \]

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

$\frac {x^n -a^n}{x-a}= (x^{n–1} + x^{n–2} a + x^{n–3} a^2+ ... + x a^{n–2} + a^{n–1}$

Hence

\[ \lim_{{x \to a}} \frac {x^n -a^n}{x-a} = a^{n – l} + aa^{n–2} +. . . + a^{n–2} (a) +a^{n–l}= na^{n-1} \]

\[ \lim_{{x \to a}} f(x) \] and \[ \lim_{{x \to a}} g(x) \]

exists, then

$\lim_{{x \to a}} f(x) \leq \lim_{{x \to a}} g(x)$

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

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

$cos x < \frac {sin x}{x} < 1$ for $ 0 < |x| < \frac {\pi}{2}$

Lets define the limit of two important Theorem based on Above Formula and Sandwitch theorem

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

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

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

Therefore from Sandwitch theorem

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

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

$\lim_{{x \to 0}} \frac {1-cosx}{x} =\lim_{{x \to 0}} \frac {2sin^2(x/2)}{x} =\lim_{{x \to 0}} \frac {sin^2(x/2)}{x/2}$

$=\lim_{{x \to 0}} \frac {sin(x/2)}{x/2} sin (x/2) = 1.0 =1$

Therefore

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

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

$\lim_{{x \to 0}} \frac {tanx}{x}=\lim_{{x \to 0}} \frac {sinx}{x cosx}= \lim_{{x \to 0}} \frac {sinx }{x} \lim_{{x \to 0}} \frac {1}{cos x}=1$

Therefore

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

(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.

**Notes****NCERT Solutions & Assignments**