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:
6.2 Rationalization
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 15.
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2 + x} - \sqrt 2 }}{x}$.
Solution:
$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2 + x} - \sqrt 2 }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2 + x} - \sqrt 2 }}{x} \times \frac{{\sqrt {2 + x} + \sqrt 2 }}{{\sqrt {2 + x} + \sqrt 2 }}$ [Rationalization of Numerator]
$ = \mathop {\lim }\limits_{x \to 0} \frac{{(2 + x) - 2}}{{x(\sqrt {2 + x} + \sqrt 2 )}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{x}{{x(\sqrt {2 + x} + \sqrt 2 )}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {2 + x} + \sqrt 2 }}$
$ = \frac{1}{{2\sqrt 2 }}$
Question 16.
Evaluate $\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } \right)$.
Solution:
$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } \right) \times \frac{{\left( {\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} } \right)}}{{\left( {\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} } \right)}}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{x^2} + x + 1} \right) - \left( {{x^2} + 1} \right)}}{{\left( {\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} } \right)}}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{x}{{\left( {\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} } \right)}}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{x}{{x\left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + \sqrt {1 + \frac{1}{{{x^2}}}} } \right)}}$
$ = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{\left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + \sqrt {1 + \frac{1}{{{x^2}}}} } \right)}}$
$ = \frac{1}{{\left( {1 + 1} \right)}}$ [as $x \to \infty $, $\frac{1}{x} \to 0$ ]
=$\frac{1}{2}$
