Page 203 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes Now let us find the limiting value of the function (5x–3) when x approaches 0.
i.e.

For finding this limit, we assign values to x from left and also from right of 0.
x –0.1 –0.01 –0.001 –0.0001..........
5x – 3 –3.5 –3.05 –3.005 –3.0005..........
x –0.1 –0.01 –0.001 –0.0001..........
5x – 3 –2.5 –2.95 –2.995 –2.9995..........
It is clear from the above that the limit of (5x–3) as x 0 is –3
= –3
i.e.,
This is illustrated graphically in the Figure 7.1.
Figure 7.1

The method of finding limiting values of a function at a given point by putting the values of the

variable very close to that point may not always be convenient.
We, therefore, need other methods for calculating the limits of a function as x (independent
variable) ends to a finite quantity.

Consider an example: Find

We can solve it by the method of substitution. Steps of which are as follows:

Step 1: We consider a value of x close to a say x For f(x) = we write x = 3 + h, so that
= a + h, where h is a very small positive
number. Clearly, as x a, h 0 as x 3, h 0

Step 2: Simplify f(x) = f(a + h) Now f(x) = f(3+h)
=

=

= h+6
Contd...

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