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:
1.2 Questions
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 1.
If $f(x) = x + 5$.
Find $\mathop {\lim }\limits_{x \to 2} f(x)$.
Solution:
Let us find the limit of $f(x)$ as $x$ approaches 2 by using the definition of Limit.
| $x$ | $f(x)$ |
| 1.98 | (1.98)+5= 6.98 |
| 1.999 | 6.999 |
| 1.99999 | 6.99999 |
| 2.0001 | 7.0001 |
| 2.02 | 7.02 |
We observe as value of $x$ approaches 2, $f(x)$ value is moving close to 7.
$\therefore $ $\mathop {\lim }\limits_{x \to 2} f(x) = \mathop {\lim }\limits_{x \to 2} (x + 5) = 7$
Question 2.
If $f(x)$=$\frac{{\sin x}}{x}$.
Find $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}$, where $x$ in radians.
Solution:
Let us find the limit of $f(x)$ as $x$ approaches 0 by using the definition of Limit.
| $x$ | $\sin x$ | $f(x) = \frac{{\sin x}}{x}$ |
| -0.0005 | -4.9999x10-4 | 0.99999 |
| -0.00001 | -10-5 | 1.00000 |
| 0.00001 | 0.00001 | 1.00000 |
| 0.0006 | 5.9999x10-4 | 0.99999 |
| 0.006 | 5.9999x10-4 | 0.999994 |
As x$ \to $0, $f(x)$ value approaches 1.
$\therefore \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$
