Limits
1.0 Introduction
1.1 Basic Method of Evaluation of Limits:
1.2 Questions
1.3 Formal definition of Limit:
1.4 Evaluation of Limits by Direct Substitution Method:
1.5 Neighbourhood Concept:
2.0 Definition of Limit - In a different form:
2.1 One - Sided Limits:
2.2 Left hand Limit of a function:(LHL)
2.3 Right hand Limit of a function:(RHL)
3.0 Conditions for existence of Limit
4.0 Some Standard Limits
5.0 Algebra of limits
6.0 Some Standard Methods of Evaluation of Limits:
7.0 Indeterminate Forms:
7.1 Limits of the form ${1^\infty }$
7.2 Limits of the form ${0^0}$
7.3 Limits of the form ${\infty^0}$
7.4 Limit of a function as $x \to \infty $ :
8.0 Sandwich Theorem / Squeeze Play Theorem:
9.0 L'Hospital's Rule for evaluation of limits:
6.1 Factorization
1.2 Questions
1.3 Formal definition of Limit:
1.4 Evaluation of Limits by Direct Substitution Method:
1.5 Neighbourhood Concept:
2.2 Left hand Limit of a function:(LHL)
2.3 Right hand Limit of a function:(RHL)
7.2 Limits of the form ${0^0}$
7.3 Limits of the form ${\infty^0}$
7.4 Limit of a function as $x \to \infty $ :
Question 13.
Evaluate $\mathop {\lim }\limits_{x \to 2} \frac{{{x^6} - 24x - 16}}{{{x^3} + 2x - 12}}$.
Solution:
$\mathop {\lim }\limits_{x \to 2} \frac{{{x^6} - 24x - 16}}{{{x^3} + 2x - 12}} = \mathop {\lim }\limits_{x \to 2} \frac{{(x - 2)({x^5} + 2{x^4} + 4{x^3} + 8{x^2} + 16x + 8)}}{{(x - 2)({x^2} + 2x + 6)}}$
$ = \mathop {\lim }\limits_{x \to 2} \frac{{({x^5} + 2{x^4} + 4{x^3} + 8{x^2} + 16x + 8)}}{{({x^2} + 2x + 6)}}$
$ = \mathop {\lim }\limits_{x \to 2} \frac{{{{(2)}^5} + 2{{(2)}^4} + 4{{(2)}^3} + 8{{(2)}^2} + 16(2) + 8}}{{{{(2)}^2} + 2(2) + 6}}$
$ = \frac{{168}}{{14}}$
$ = 12$
Question 14.
Evaluate $\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}}$.
Solution:
$\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} = \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}$
$ = \mathop {\lim }\limits_{x \to 9} \frac{1}{{\left( {\sqrt x + 3} \right)}}$
$ = \frac{1}{{\sqrt 9 + 3}}$
$ = \frac{1}{6}$
