Gujarat Board Statistics Class 12 GSEB Solutions Part 2 Chapter 4 Limit Ex 4.2 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4.2

Question 1.

Find the values of the following using tabular method:

(1)

Answer:

Here, f(x) = 2x + 1. Taking the values of x very close to 1, the following table is prepared:

It is clear from the table that when x is brought nearer to 1 by increasing or decreasing its value, the value of f(x) approaches to 3. That is, when x â†’ 1, f(x) â†’ 3

(2)

Answer:

Here, f(x) = \(\frac{x^{2}-2 x-3}{x-3}\)

= \(\frac{x^{2}-3 x+x-3}{x-3}\)

= \(\frac{(x+1)(x-3)}{x-3}\)

= x + 1

Taking the values of x very close to 3, the following table is prepared:

It is clear from the table that when x is brought nearer to 3 by increasing or decreasing its value, the value of f(x) approaches to 4. That is, when x â†’ 3, f(x) â†’ 4.

(3)

Answer:

Here, f(x) = \(\frac{2 x^{2}+3 x-14}{x-2}\)

= \(\frac{2 x^{2}+7 x-4 x-14}{x-2}\)

= \(\frac{(x-2)(2 x+7)}{x-2}\)

= 2x + 7

Taking the values of x very close to 2, the following table is prepared:

It is clear from the table that when x is brought nearer to 2 by increasing or decreasing its value, the value of f(x) approaches to 11. That is, when x â†’ 2, f(x) â†’ 11

(4)

Answer:

Here, f(x) = \(\frac{2 x^{2}+9 x+9}{x+3}\)

= \(\frac{2 x^{2}+6 x+3 x+9}{x+3}\)

= \(\frac{(2 x+3)(x+3)}{x+3}\)

= 2x + 3

Taking the values of x very close to – 3, the following table is prepared:

It is clear from the table that when x is brought nearer to – 3 by increasing or decreasing its value, the value of f(x) approaches to – 3. That is, when x â†’ – 3, f(x) â†’ – 3

(5)

Answer:

Here, f(x) = x. Taking the values of x very close to 2, the following table is prepared:

It is clear from the table that when x is brought nearer to 2 by increasing or decreasing its value, the value of f(x) approaches to 2. That is, when x â†’ 2, f(x) â†’ 2.

Question 2.

Using tabular method, show that does not exist.

Answer:

Here, f(x) = \(\frac{2}{x-3}\). Taking the values of x very close to 3. the following table is prepared:

It is clear from the table that when x is brought nearer to 3 by increasing or decreasing its value, the value of f(x) does not approaches to a definite value. That is, when x â†’ 3, f(x) does not tend to a definite value. Hence, the limit of this function does not exist.

does not exist.

Question 3.

If y = \(\frac{x^{2}+x-6}{x-2}\), show that as x â†’ 2, then y â†’ 5 using tabular method.

Answer:

Here, y = \(\frac{x^{2}+x-6}{x-2}\)

= \(\frac{x^{2}+3 x-2 x-6}{x-2}\)

= \(\frac{(x-2)(x+3)}{x-2}\)

= x + 3

We have to prove that, when x â†’ 2, then y â†’ 5. So we have to find

Here, y = x + 3. Taking the values of x very close to 2, the following table is prepared :

It is clear from the table that when x is brought nearer to 2 by increasing or decreasing its value, the value of y approaches to 5. That is, when x â†’ 2, y â†’ 5.

Hence, it is proved that when x â†’ 2, then y â†’ 5.

Question 4.

If y = 5 – 2x, show that as x â†’ – 1 then y â†’ 7 using tabular method.

Answer:

Here, y = 5 – 2x. Taking the values of x very close to – 1, the following table is prepared:

It is clear from the table that when x is brought nearer to – 1 by increasing or decreasing its value, the value of y approaches to 7. That is, when x â†’ – 1, y â†’ 7.

Hence, it is proved that when x â†’ – 1, then y â†’ 7.