Open in App
Not now

# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.4 | Set 2

• Last Updated : 28 Apr, 2021

### Question 18. Limxâ†’1{âˆš(5x – 4) – âˆšx}/(x3 – 1)

Solution:

We have, Limxâ†’1{âˆš(5x – 4) – âˆšx}/(x3 – 1)

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’1{(5x – 4) – x}/[{âˆš(5x – 4) + âˆšx}(x3 – 1)]

= Limxâ†’1{4(x – 1)}/[{âˆš(5x – 4) + âˆšx}(x-1)(x2 + x + 1)]

= Limxâ†’1(4)/[{âˆš(5x – 4) + âˆšx}(x2 + x + 1)]

Now put x = 1, we get

= 4/{(3)(âˆš1 + âˆš1)}

= 4/6

= 2/3

### Question 19. Limxâ†’2{âˆš(1 + 4x) – âˆš(5 + 2x)}/(x – 2)

Solution:

We have, Limxâ†’2{âˆš(1 + 4x) – âˆš(5 + 2x)}/(x – 2)

Find the limit of the given equation

When we put x = 2, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’2{âˆš(1 + 4x) – âˆš(5+2x)}/[(x – 2){âˆš(1 + 4x) + âˆš(5 + 2x)}]

= Limxâ†’2{(1 + 4x) – (5 + 2x)}/[(x – 2){âˆš(1 + 4x) + âˆš(5 + 2x)}]

= Limxâ†’2{2(x – 2)}/[(x – 2){âˆš(1 + 4x) + âˆš(5 + 2x)}]

= Limxâ†’2(2)/{âˆš(1 + 4x) + âˆš(5 + 2x)}

Now put x = 2, we get

= 2/{âˆš(1 + 8) + âˆš(5 + 4)}

= 2/(3 + 3)

= 1/3

### Question 20. Limxâ†’1{âˆš(3 + x) – âˆš(5 – x)}/(x2 – 1)

Solution:

We have, Limxâ†’1{âˆš(3 + x) – âˆš(5 – x)}/(x2 – 1)

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’1{(3 + x) – (5 – x)}/[(x2 – 1){âˆš(3 + x) + âˆš(5 – x)}]

= Limxâ†’1{2(x – 1)}/[(x – 1)(x + 1){âˆš(3 + x) + âˆš(5 – x)}]

= Limxâ†’1(2)/[(x + 1){âˆš(3 + x) + âˆš(5 – x)}]

Now put x = 1, we get

= 2/{2(2 + 2)}

= 1/4

### Question 21. Limxâ†’0{âˆš(1 + x2) – âˆš(1 – x2)}/(x)

Solution:

We have, Limxâ†’0{âˆš(1 + x2) – âˆš(1 – x2)}/(x)

Find the limit of the given equation

When we put x = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’0{(1 + x2) – (1 – x2)}/[x{âˆš(1 + x2) + âˆš(1 – x2)}]

= Limxâ†’0{(1 + x2) – (1 – x2)}/[x{âˆš(1 + x2) + âˆš(1 – x2)}]

= Limxâ†’0(2x2/[x{âˆš(1 + x2) + âˆš(1 – x2)}]

= Limxâ†’0(2x/{âˆš(1 + x2) + âˆš(1 – x2)}

Now put x = 0, we get

= 2 Ã— 0/(âˆš1 + âˆš1)

= 0

### Question 22. Limxâ†’0{âˆš(1 + x + x2) – âˆš(x + 1)}/(2x2)

Solution:

We have, Limxâ†’0{âˆš(1 + x + x2) – âˆš(x + 1)}/(2x2)

Find the limit of the given equation

When we put x = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’0{(1 + x + x2) – (x + 1)}/[2x2{âˆš(1 + x + x2) – âˆš(x + 1)}]

= Limxâ†’0(x2)/[2x2{âˆš(1 + x + x2) – âˆš(x + 1)}]

= Limxâ†’0(1)/[2{âˆš(1 + x + x2) – âˆš(x + 1)}]

Now put x = 0, we get

= 1/{2(âˆš1 + âˆš1)

= 1/4

### Question 23. Limxâ†’4{2 – âˆšx}/(4 – x)

Solution:

We have, Limxâ†’4{2 – âˆšx}/(4 – x)

Find the limit of the given equation

When we put x = 4, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’4{4 – x}/[(4 – x){2 + âˆšx}]

= Limxâ†’4{4 – x}/[(4 – x){2 + âˆšx}]

= Limxâ†’4(1)/{2 + âˆšx}

Now put x = 4, we get

= 1/(2 + 2)

= 1/4

### Question 24. Limxâ†’a(x – a)/{âˆšx – âˆša}

Solution:

We have, Limxâ†’a(x – a)/{âˆšx – âˆša}

Find the limit of the given equation

When we put x = a, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

=

= Limxâ†’a[(x – a){âˆšx – âˆša}]/(x – a)

= Limxâ†’a{âˆšx + âˆša}

Now put x = a, we get

= âˆša + âˆša

= 2âˆša

### Question 25. Limxâ†’0{âˆš(1 + 3x) – âˆš(1 – 3x)}/(x)

Solution:

We have, Limxâ†’0{âˆš(1 + 3x) – âˆš(1 – 3x)}/(x)

Find the limit of the given equation

When we put x = a, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’0{(1 + 3x) – (1 – 3x)}/[(x){âˆš(1 + 3x) – âˆš(1 – 3x)}]

= Limxâ†’0(6x)/[(x){âˆš(1 + 3x) – âˆš(1 – 3x)}]

= Limxâ†’0(6)/{âˆš(1 + 3x) – âˆš(1 – 3x)}

Now put x = 0, we get

= 6/(âˆš1 + âˆš1)

= 6/2

= 3

### Question 26. Limxâ†’0{âˆš(2 – x) – âˆš(2 + x)}/(x)

Solution:

We have, Limxâ†’0{âˆš(2 – x) – âˆš(2 + x)}/(x)

Find the limit of the given equation

When we put x = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’0{(2 – x) – (2 + x)}/[x{âˆš(2 – x) + âˆš(2 + x)}]

= Limxâ†’0(-2x)/[x{âˆš(2 – x) + âˆš(2 + x)}]

= Limxâ†’0(-2)/{âˆš(2 – x) + âˆš(2 + x)}

Now put x = 0, we get

= (-2)/(âˆš2 + âˆš2)

= (-2)/(2âˆš2)

= -1/(âˆš2)

### Question 27. Limxâ†’1{âˆš(3 + x) – âˆš(5 – x)}/(x2 – 1)

Solution:

We have, Limxâ†’1{âˆš(3 + x) – âˆš(5 – x)}/(x2 – 1)

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’1{(3 + x) – (5 – x)}/[(x2 – 1){âˆš(3 + x) + âˆš(5 – x)}]

= Limxâ†’1{2(x – 1)}/[(x – 1)(x + 1){âˆš(3 + x) + âˆš(5 – x)}]

= Limxâ†’1(2)/[(x + 1){âˆš(3 + x) + âˆš(5 – x)}]

Now put x = 1, we get

= 2/{(2)(âˆš4 + âˆš4)}

= 2/8

= 1/4

### Question 28. Limxâ†’1{(2x – 3)(âˆšx – 1)}/(3x2 + 3x – 6)

Solution:

We have, Limxâ†’1{(2x – 3)(âˆšx – 1)}/(3x2 + 3x – 6)

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’1{(2x – 3)(x – 1)}/[(3x2 + 3x – 6)(âˆšx + 1)]

= Limxâ†’1{(2x – 3)(x – 1)}/[3(x2 + x – 2)(âˆšx + 1)]

= Limxâ†’1{(2x – 3)(x – 1)}/[3(x – 1)(x + 2)(âˆšx + 1)]

= Limxâ†’1(2x – 3)/[3(x + 2)(âˆšx + 1)]

Now put x = 1, we get

= (2 – 3)/{3(3)(âˆš1 + 1)

= -1/(3 Ã— 3 Ã— 2)

= -1/18

### Question 29. Limxâ†’0{âˆš(1 + x2) – âˆš(1 + x)}/{âˆš(1 + x3) – âˆš(1 + x)}

Solution:

We have, Limxâ†’0{âˆš(1 + x2) – âˆš(1 + x)}/{âˆš(1 + x3) – âˆš(1 + x)}

Find the limit of the given equation

When we put x = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

Now put x = 0, we get

= (âˆš1 + âˆš1)/{1(âˆš1 + âˆš1)}

= 2/2

= 1

### Question 30. Limxâ†’1{x2 – âˆšx}/{âˆšx – 1}

Solution:

We have, Limxâ†’1{x2 – âˆšx}/{âˆšx – 1}

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’1{âˆšx(xâˆšx -1)}/{âˆšx – 1}

= Limxâ†’1{âˆšx(x3/2 – 1)}/{âˆšx – 1}

= Limxâ†’1[âˆšx{(âˆšx)3 – 1}]/{âˆšx – 1}

= Limxâ†’1[(âˆšx)(âˆšx – 1)(x + âˆšx + 1)]/{âˆšx – 1}

= Limxâ†’1[(âˆšx)(x + âˆšx + 1)]

Now put x = 1, we get

= (âˆš1)(1 + âˆš1 + 1)

= 3

### Question 31. Limhâ†’0{âˆš(x + h) – âˆšx}/(h), x â‰  0

Solution:

We have, Limhâ†’0{âˆš(x + h) – âˆšx}/(h)

Find the limit of the given equation

When we put h = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limhâ†’0{(x + h) – x}/[h{âˆš(x + h) + âˆšx}]

= Limhâ†’0(h)/[h{âˆš(x + h) + âˆšx}]

= Limhâ†’0(1)/{âˆš(x + h) + âˆšx}

Now put x = 0, we get

= 1/(âˆšx + âˆšx)

= 1/(2âˆšx)

### Question 32. Limxâ†’âˆš10{âˆš(7 + 2x) – (âˆš5 + âˆš2)}/(x2 – 10)

Solution:

We have, Limxâ†’âˆš10{âˆš(7 + 2x) – (âˆš5 + âˆš2)}/(x2 – 10)

= Limxâ†’âˆš10{âˆš(7 + 2x) – âˆš(âˆš5 + âˆš2)2}/{(x – âˆš10)(x + âˆš10)}

= Limxâ†’âˆš10{âˆš(7 + 2x) – âˆš(5 + 2 + 2âˆš5âˆš2)}/{(x – âˆš10)(x + âˆš10)}

= Limxâ†’âˆš10{âˆš(7 + 2x) – âˆš(7 + 2âˆš10)}/{(x – âˆš10)(x + âˆš10)}

On rationalizing numerator.

=

=

=

Now put x = âˆš10, we get

=

=

=

= 1/{(2âˆš10)(âˆš5 + âˆš2)}

On rationalizing denominator.

= (âˆš5 – âˆš2)/{(2âˆš10)(5 – 2)}

= (âˆš5 – âˆš2)/(6âˆš10)

### Question 33. Limxâ†’âˆš6{âˆš(5 + 2x) – (âˆš3 + âˆš2)}/(x2 – 6)

Solution:

We have, Limxâ†’âˆš6{âˆš(5 + 2x) – (âˆš3 + âˆš2)}/(x2 – 6)

= Limxâ†’âˆš6{âˆš(5 + 2x) – âˆš(âˆš3 + âˆš2)2}/{(x – âˆš6)(x + âˆš6)}

= Limxâ†’âˆš6{âˆš(5 + 2x) – âˆš(3 + 2 + 2âˆš3âˆš2)}/{(x – âˆš6)(x + âˆš6)}

= Limxâ†’âˆš6{âˆš(5 + 2x) – âˆš(5 + 2âˆš6)}/{(x  -âˆš6)(x + âˆš6)}

On rationalizing numerator.

=

=

=

Now put x = âˆš6, we get

=

=

=

= 1/{(2âˆš6)(âˆš3 + âˆš2)}

On rationalizing denominator, we get

= (âˆš3 – âˆš2)/{(2âˆš6)(3 – 2)}

= (âˆš3 – âˆš2)/(2âˆš6)

### Question 34. Limxâ†’âˆš2{âˆš(3 + 2x) – (âˆš2 + 1)}/(x2 – 2)

Solution:

We have,  Limxâ†’âˆš2{âˆš(3 + 2x) – (âˆš2 + 1)}/(x2 – 2)

= Limxâ†’âˆš2{âˆš(3 + 2x) – âˆš(âˆš2 + 1)2}/{(x – âˆš2)(x + âˆš2)}

= Limxâ†’âˆš2{âˆš(3 + 2x) – âˆš(2 + 1 + 2âˆš3)}/{(x – âˆš2)(x + âˆš2)}

= Limxâ†’âˆš2{âˆš(3 + 2x) – âˆš(3 + 2âˆš3)}/{(x – âˆš2)(x + âˆš2)}

On rationalizing numerator.

=

=

=

=

=

Now put x = âˆš2, we get

=

=

=

= 1/{(2âˆš2)(âˆš2 + 1)}

On rationalizing denominator, we get

= (âˆš2 – 1)/{(2âˆš2)(2 – 1)}

= (âˆš2 – 1)/(2âˆš2)

My Personal Notes arrow_drop_up
Related Articles