Skip to content
Related Articles
Get the best out of our app
GeeksforGeeks App
Open App
geeksforgeeks
Browser
Continue

Related Articles

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

Improve Article
Save Article
Like Article
author
vivekray59
proficient
41 published articles
Improve Article
Save Article
Like Article

Evaluates the following limits:

Question 18. Limx→1{√(5x – 4) – √x}/(x3 – 1)

Solution:

We have, Limx→1{√(5x – 4) – √x}/(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

= Lim_{x→1}\frac{\sqrt{(5x - 4)} - \sqrt{x}}{(x^3 - 1)} \times \frac{\sqrt{(5x - 4)} + \sqrt{x}}{\sqrt{(5x - 4)} + \sqrt{x}}

= Limx→1{(5x – 4) – x}/[{√(5x – 4) + √x}(x3 – 1)]

= Limx→1{4(x – 1)}/[{√(5x – 4) + √x}(x-1)(x2 + x + 1)]

= Limx→1(4)/[{√(5x – 4) + √x}(x2 + x + 1)]

Now put x = 1, we get

= 4/{(3)(√1 + √1)}

= 4/6

= 2/3

Question 19. Limx→2{√(1 + 4x) – √(5 + 2x)}/(x – 2)

Solution:

We have, Limx→2{√(1 + 4x) – √(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

= Lim_{x→2}\frac{\sqrt{(1 + 4x)} - \sqrt{(5 + 2x)}}{(x - 2)} \times \frac{\sqrt{(1 + 4x)} + \sqrt{(5 + 2x)}}{\sqrt{(1 + 4x)} + \sqrt{(5 + 2x)}}

= Limx→2{√(1 + 4x) – √(5+2x)}/[(x – 2){√(1 + 4x) + √(5 + 2x)}]

= Limx→2{(1 + 4x) – (5 + 2x)}/[(x – 2){√(1 + 4x) + √(5 + 2x)}]

= Limx→2{2(x – 2)}/[(x – 2){√(1 + 4x) + √(5 + 2x)}]

= Limx→2(2)/{√(1 + 4x) + √(5 + 2x)}

Now put x = 2, we get

= 2/{√(1 + 8) + √(5 + 4)}

= 2/(3 + 3)

= 1/3

Question 20. Limx→1{√(3 + x) – √(5 – x)}/(x2 – 1)

Solution:

We have, Limx→1{√(3 + x) – √(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

= Lim_{x→1}\frac{\sqrt{(3 + x)} - \sqrt{(5 - x)}}{(x^2 - 1)} \times \frac{\sqrt{(3 + x)} + \sqrt{(5 - x)}}{\sqrt{(3 + x)} + \sqrt{(5 - x)}}

= Limx→1{(3 + x) – (5 – x)}/[(x2 – 1){√(3 + x) + √(5 – x)}]

= Limx→1{2(x – 1)}/[(x – 1)(x + 1){√(3 + x) + √(5 – x)}]

= Limx→1(2)/[(x + 1){√(3 + x) + √(5 – x)}]

Now put x = 1, we get

= 2/{2(2 + 2)}

= 1/4

Question 21. Limx→0{√(1 + x2) – √(1 – x2)}/(x)

Solution:

We have, Limx→0{√(1 + x2) – √(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

= Lim_{x→0}\frac{\sqrt{(1 + x^2)} - \sqrt{(1 - x^2)}}{x} \times \frac{\sqrt{(1 + x^2)} + \sqrt{(1 - x^2)}}{\sqrt{(1 + x^2)} + \sqrt{(1 - x^2)}}

= Limx→0{(1 + x2) – (1 – x2)}/[x{√(1 + x2) + √(1 – x2)}]

= Limx→0{(1 + x2) – (1 – x2)}/[x{√(1 + x2) + √(1 – x2)}]

= Limx→0(2x2/[x{√(1 + x2) + √(1 – x2)}]

= Limx→0(2x/{√(1 + x2) + √(1 – x2)}

Now put x = 0, we get

= 2 × 0/(√1 + √1)

= 0

Question 22. Limx→0{√(1 + x + x2) – √(x + 1)}/(2x2)

Solution:

We have, Limx→0{√(1 + x + x2) – √(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

= Lim_{x→0}\frac{\sqrt{(1 + x + x^2)} - \sqrt{(x + 1)}}{2x^2} \times \frac{\sqrt{(1 + x + x^2)} + \sqrt{(x + 1)}}{\sqrt{(1 + x + x^2)} + \sqrt{(x + 1)}}

= Limx→0{(1 + x + x2) – (x + 1)}/[2x2{√(1 + x + x2) – √(x + 1)}]

= Limx→0(x2)/[2x2{√(1 + x + x2) – √(x + 1)}]

= Limx→0(1)/[2{√(1 + x + x2) – √(x + 1)}]

Now put x = 0, we get

= 1/{2(√1 + √1)

= 1/4

Question 23. Limx→4{2 – √x}/(4 – x)

Solution:

We have, Limx→4{2 – √x}/(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

= Lim_{x→4}\frac{2 - √x}{(4 - x)} \times \frac{2 + √x}{2 + √x}

= Limx→4{4 – x}/[(4 – x){2 + √x}]

= Limx→4{4 – x}/[(4 – x){2 + √x}]

= Limx→4(1)/{2 + √x}

Now put x = 4, we get

= 1/(2 + 2)

= 1/4

Question 24. Limx→a(x – a)/{√x – √a}

Solution:

We have, Limx→a(x – a)/{√x – √a}

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

=   Lim_{x→a}\frac{(x - a)}{√x - √a} \times \frac{√x + √a}{√x + √a}

= Limx→a[(x – a){√x – √a}]/(x – a)

= Limx→a{√x + √a}

Now put x = a, we get

= √a + √a

= 2√a

Question 25. Limx→0{√(1 + 3x) – √(1 – 3x)}/(x)

Solution:

We have, Limx→0{√(1 + 3x) – √(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

= Lim_{x→0}\frac{\sqrt{(1 + 3x)} - \sqrt{(1 - 3x)}}{x} \times \frac{\sqrt{(1 + 3x)} + \sqrt{(1 - 3x)}}{\sqrt{(1 + 3x)} + \sqrt{(1 - 3x)}}

= Limx→0{(1 + 3x) – (1 – 3x)}/[(x){√(1 + 3x) – √(1 – 3x)}]

= Limx→0(6x)/[(x){√(1 + 3x) – √(1 – 3x)}]

= Limx→0(6)/{√(1 + 3x) – √(1 – 3x)}

Now put x = 0, we get

= 6/(√1 + √1)

= 6/2

= 3

Question 26. Limx→0{√(2 – x) – √(2 + x)}/(x)

Solution:

We have, Limx→0{√(2 – x) – √(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

=  Lim_{x→0}\frac{\sqrt{(2 - x)} - \sqrt{(2 + x)}}{x} \times \frac{\sqrt{(2 - x)} + \sqrt{(2 + x)}}{\sqrt{(2 - x)} + \sqrt{(2 + x)}}

= Limx→0{(2 – x) – (2 + x)}/[x{√(2 – x) + √(2 + x)}]

= Limx→0(-2x)/[x{√(2 – x) + √(2 + x)}]

= Limx→0(-2)/{√(2 – x) + √(2 + x)}

Now put x = 0, we get

= (-2)/(√2 + √2)

= (-2)/(2√2)

= -1/(√2)

Question 27. Limx→1{√(3 + x) – √(5 – x)}/(x2 – 1)

Solution:

We have, Limx→1{√(3 + x) – √(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

= Lim_{x→1}\frac{\sqrt{(3 + x)} - \sqrt{(5 - x)}}{(x^2 - 1)} \times \frac{\sqrt{(3 + x)} + \sqrt{(5 - x)}}{\sqrt{(3 + x)} + \sqrt{(5 - x)}}

= Limx→1{(3 + x) – (5 – x)}/[(x2 – 1){√(3 + x) + √(5 – x)}]

= Limx→1{2(x – 1)}/[(x – 1)(x + 1){√(3 + x) + √(5 – x)}]

= Limx→1(2)/[(x + 1){√(3 + x) + √(5 – x)}]

Now put x = 1, we get

= 2/{(2)(√4 + √4)}

= 2/8

= 1/4

Question 28. Limx→1{(2x – 3)(√x – 1)}/(3x2 + 3x – 6)

Solution:

We have, Limx→1{(2x – 3)(√x – 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

= Limx→1{(2x – 3)(x – 1)}/[(3x2 + 3x – 6)(√x + 1)]

= Limx→1{(2x – 3)(x – 1)}/[3(x2 + x – 2)(√x + 1)]

= Limx→1{(2x – 3)(x – 1)}/[3(x – 1)(x + 2)(√x + 1)]

= Limx→1(2x – 3)/[3(x + 2)(√x + 1)]

Now put x = 1, we get

= (2 – 3)/{3(3)(√1 + 1)

= -1/(3 × 3 × 2)

= -1/18

Question 29. Limx→0{√(1 + x2) – √(1 + x)}/{√(1 + x3) – √(1 + x)}

Solution:

We have, Limx→0{√(1 + x2) – √(1 + x)}/{√(1 + x3) – √(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

=\lim_{x\to0}\frac{(\sqrt{1+x^2}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})}{(\sqrt{1+x^3}-\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})}

=\lim_{x\to0}\frac{x(x-1)}{x(x^2-1)}×\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}

=\lim_{x\to0}\frac{[(1+x^2)-(1+x)]}{[(1+x^3)-(1+x)}×\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}

=\lim_{x\to0}\frac{(x^2-x)}{(x^3-x)}×\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}

=\lim_{x\to0}\frac{1}{(x+1)}×\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}

Now put x = 0, we get

= (√1 + √1)/{1(√1 + √1)}

= 2/2

= 1

Question 30. Limx→1{x2 – √x}/{√x – 1}

Solution:

We have, Limx→1{x2 – √x}/{√x – 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

= Limx→1{√x(x√x -1)}/{√x – 1}

= Limx→1{√x(x3/2 – 1)}/{√x – 1}

= Limx→1[√x{(√x)3 – 1}]/{√x – 1}

= Limx→1[(√x)(√x – 1)(x + √x + 1)]/{√x – 1}

= Limx→1[(√x)(x + √x + 1)]

Now put x = 1, we get

= (√1)(1 + √1 + 1)

= 3

Question 31. Limh→0{√(x + h) – √x}/(h), x ≠ 0   

Solution:

We have, Limh→0{√(x + h) – √x}/(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

= Lim_{h→0}\frac{\sqrt{(x + h)} - \sqrt{x}}{h} \times \frac{\sqrt{(x + h)} + \sqrt{x}}{\sqrt{(x + h)} + \sqrt{x}}

= Limh→0{(x + h) – x}/[h{√(x + h) + √x}]

= Limh→0(h)/[h{√(x + h) + √x}]

= Limh→0(1)/{√(x + h) + √x}

Now put x = 0, we get

= 1/(√x + √x)

= 1/(2√x)

Question 32. Limx→√10{√(7 + 2x) – (√5 + √2)}/(x2 – 10)

Solution:

We have, Limx→√10{√(7 + 2x) – (√5 + √2)}/(x2 – 10)

= Limx→√10{√(7 + 2x) – √(√5 + √2)2}/{(x – √10)(x + √10)}

= Limx→√10{√(7 + 2x) – √(5 + 2 + 2√5√2)}/{(x – √10)(x + √10)}

= Limx→√10{√(7 + 2x) – √(7 + 2√10)}/{(x – √10)(x + √10)}

On rationalizing numerator.

=\lim_{x\to\sqrt{10}}\frac{(\sqrt{7+2x}-\sqrt{7+2\sqrt{10}})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}{(x-\sqrt{10})(x+\sqrt{10})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}

=\lim_{x\to\sqrt{10}}\frac{(7+2x)-(7+2\sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}

=\lim_{x\to\sqrt{10}}\frac{2(x-\sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}

Now put x = √10, we get

=\frac{2}{(\sqrt{10}+\sqrt{10})(2\sqrt{7+2\sqrt{10}}}

=\frac{1}{(2\sqrt{10})(\sqrt{7+2\sqrt{10}}}

=\frac{1}{(2\sqrt{10})(\sqrt{\sqrt{5}+\sqrt{2})^2}}

= 1/{(2√10)(√5 + √2)}

On rationalizing denominator.

= (√5 – √2)/{(2√10)(5 – 2)}

= (√5 – √2)/(6√10)

Question 33. Limx→√6{√(5 + 2x) – (√3 + √2)}/(x2 – 6)

Solution:

We have, Limx→√6{√(5 + 2x) – (√3 + √2)}/(x2 – 6)

= Limx→√6{√(5 + 2x) – √(√3 + √2)2}/{(x – √6)(x + √6)}

= Limx→√6{√(5 + 2x) – √(3 + 2 + 2√3√2)}/{(x – √6)(x + √6)}

= Limx→√6{√(5 + 2x) – √(5 + 2√6)}/{(x  -√6)(x + √6)}

On rationalizing numerator.

= \lim_{x\to\sqrt{6}}\frac{(\sqrt{5+2x}-\sqrt{5+2\sqrt{6}})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}

=\lim_{x\to\sqrt{6}}\frac{(5+2x)-(5+2\sqrt{6})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}

=\lim_{x\to\sqrt{6}}\frac{2(x-\sqrt{6})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}

=\lim_{x\to\sqrt{6}}\frac2{(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}

Now put x = √6, we get

=\frac2{(\sqrt6+\sqrt{6})(\sqrt{5+2\sqrt{5}}+\sqrt{5+2\sqrt{6}})}

=\frac2{(2\sqrt{6})(2\sqrt{5+2\sqrt{5}})}

=\frac1{(2\sqrt{6})(\sqrt{5+2\sqrt{5}})}

= 1/{(2√6)(√3 + √2)}

On rationalizing denominator, we get

= (√3 – √2)/{(2√6)(3 – 2)}

= (√3 – √2)/(2√6)

Question 34. Limx→√2{√(3 + 2x) – (√2 + 1)}/(x2 – 2)

Solution:

We have,  Limx→√2{√(3 + 2x) – (√2 + 1)}/(x2 – 2)

= Limx→√2{√(3 + 2x) – √(√2 + 1)2}/{(x – √2)(x + √2)}

= Limx→√2{√(3 + 2x) – √(2 + 1 + 2√3)}/{(x – √2)(x + √2)}

= Limx→√2{√(3 + 2x) – √(3 + 2√3)}/{(x – √2)(x + √2)}

On rationalizing numerator.

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

=\lim_{x\to\sqrt{2}}\frac{(3+2x)-(3+2\sqrt2)}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt2})}

=\lim_{x\to\sqrt{2}}\frac{(3+2x)-(3+2\sqrt2)}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt2})}

=\lim_{x\to\sqrt{2}}\frac{2(x-\sqrt{3})}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt2})}

=\lim_{x\to\sqrt{2}}\frac{2}{(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt2})}

Now put x = √2, we get

=\frac2{(\sqrt2+\sqrt2)(\sqrt{3+2\sqrt{5}}+\sqrt{3+2\sqrt2})}

=\frac2{(2\sqrt2)(2\sqrt{3+2\sqrt2})}

=\frac1{(2\sqrt{2})(\sqrt{3+2\sqrt2})}

= 1/{(2√2)(√2 + 1)}

On rationalizing denominator, we get

= (√2 – 1)/{(2√2)(2 – 1)}

= (√2 – 1)/(2√2)


My Personal Notes arrow_drop_up
Last Updated : 28 Apr, 2021
Like Article
Save Article
Similar Reads
1. Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.10 | Set 2
2. Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.5 | Set 1
3. Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.5 | Set 2
4. Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.10 | Set 3
5. Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.4 | Set 1
6. Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.6 | Set 1
7. Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.6 | Set 2
8. Class 11 RD Sharma solutions - Chapter 29 Limits - Exercise 29.7 | Set 1
9. Class 11 RD Sharma solutions - Chapter 29 Limits - Exercise 29.7 | Set 2
10. Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.10 | Set 1
Related Tutorials
1. Linear Algebra
2. Maths
3. Physics
4. Chemistry
5. CBSE Class 12 Syllabus 2023-2024
Previous
Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.4 | Set 1
Next
Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.5 | Set 1
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
vivekray59
vivekray59
Vote for difficulty
Article Tags :
Report Issue
We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox