ISC (Arts) Class 11CISCE
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# Limits - Limits of Polynomials and Rational Functions

#### notes

A function f is said to be a polynomial function of degree n f(x) = a_0+ a_1x+a_2x^2+ . . + a_nx^n  , where a_1s are real numbers such that a _n  ≠ 0 for some natural number n.
lim_(x->a) x = a .
Hence
lim_(x -> a)x^2 = lim_(x->a) (x.x) = lim_(x->a) x . lim_(x->a) x = a.a =a^2

An easy exercise in induction on n tells us that
lim_(x-> a) x^n = a^n
Now, let f(x) = a_0 + a_1x + a_2x^2 + ...+a_nx^n be a polynomial function.
Suppose of each of a_0 , a_1x , a_2x^2 , ...., a_nx^n  as a function , we have

lim_(x ->a) f(x) = lim_(x -> a) [a_0 + a_1x + a_2 x^2 + ...+a_nx^n]

= lim_(x -> a) a_0 + lim_(x -> a) a_1x  + lim_(x -> a) a_2x^2 + ... + a_nx^n

= a_0 +  a_1 lim_(x ->a) x + a_2 lim_(x ->a) x^2 + ... + a_n lim_(x ->a) x^n.

= a_0 + a_1a + a_2a^2 + ... + a_na^n

= f(a)
A function f is said to be a rational function, if f(x) = g(x)/(h(x)) ,  where g(x) and h(x) are polynomials such that h(x) ≠ 0.
Then lim_(x ->a) f(x) = lim_(x ->a)g(x)/(h(x)) =(lim_(x -> a) g(x))/(lim_(x ->a) h(x)) = g(a)/(h(a)).

Case 1 - h(a) = 0  and g(a) = k

g(a)/(h(a)) = k/0 = ∞
Limit does not exist (undefined)
Example -
lim_(x->2) (x^3 - 2 )/(x - 2) = (2^3 - 2)/(2-2) = (8-2)/0 = 6/0 = ∞

Case 2 -
h(a) = 0 and  g(a) = 0

g(a)/(h(a)) = 0/0
Example - lim_(x->2) (x^2 - 4)/(x-2) = (2^2 - 4)/(2 - 2) = (4 - 4)/(2 - 2) = 0/0

Case 3 -  h(a) = k  and g(a) = 0
g(a)/(h(a)) = 0/k = 0
Example - lim_(x->2) (x - 2)/(2x + 2) = (2 - 2)/(2 . 2 + 2) = 0/(4 + 2) = 0/6 = 0

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Limits of polynomial function [00:09:23]
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