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:
7.4 Limit of a function as $x \to \infty $ :
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 19.
Evaluate $\mathop {\lim }\limits_{x \to \infty } {\tan ^{ - 1}}x$.
Solution:
As $x \to \infty $, ${\tan ^{ - 1}}x \to \frac{\pi }{2}$.
$\therefore \mathop {\lim }\limits_{x \to \infty } {\tan ^{ - 1}}x = \frac{\pi }{2}$
Question 20.
Evaluate $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + 5}}{{{x^2} + 4x + 3}}$.
Solution:
As $x \to \infty $, $\frac{1}{x} \to 0$
$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + 5}}{{{x^2} + 4x + 3}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\left( {1 + \frac{5}{{{x^2}}}} \right)}}{{{x^2}\left( {1 + \frac{4}{x} + \frac{3}{{{x^2}}}} \right)}}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {1 + \frac{5}{{{x^2}}}} \right)}}{{\left( {1 + \frac{4}{x} + \frac{3}{{{x^2}}}} \right)}}$
$ = \frac{{1 + 0}}{{1 + 0 + 0}}$
$ = 1$
Question 21.
Evaluate $\mathop {\lim }\limits_{x \to \infty } \left( {x - \sqrt {{x^2} + x} } \right)$.
Solution:
$\mathop {\lim }\limits_{x \to \infty } \left( {x - \sqrt {{x^2} + x} } \right) = \mathop {\lim }\limits_{x \to \infty } \left( {x - \sqrt {{x^2} + x} } \right) \times \frac{{\left( {x + \sqrt {{x^2} + x} } \right)}}{{\left( {x + \sqrt {{x^2} + x} } \right)}}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - ({x^2} + x)}}{{x + \sqrt {{x^2} + x} }}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{ - x}}{{x + \sqrt {{x^2} + x} }}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{ - x}}{{x\left( {1 + \sqrt {1 + \frac{1}{x}} } \right)}}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 1}}{{1 + \sqrt {1 + \frac{1}{x}} }}$
$ = - \frac{1}{2}$
