⚡ Match Highlights: maths/class rd sharma solutions chapter limits exercise set

Evaluates the following limits:

Question 18. Lim_x→1{√(5x - 4) - √x}/(x³- 1)

Solution:

We have, Lim_x→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}}

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

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

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

Now put x = 1, we get

= 4/{(3)(√1 + √1)}

= 4/6

= 2/3

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

Solution:

We have, Lim_x→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)}}

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

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

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

= Lim_x→2(2)/{√(1 + 4x) + √(5 + 2x)}

Now put x = 2, we get

= 2/{√(1 + 8) + √(5 + 4)}

= 2/(3 + 3)

= 1/3

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

Solution:

We have, Lim_x→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)}}

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

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

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

Now put x = 1, we get

= 2/{2(2 + 2)}

= 1/4

Question 21. Lim_x→0{√(1 + x²) - √(1 - x²)}/(x)

Solution:

We have, Lim_x→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)}}

= Lim_x→0{(1 + x²) - (1 - x²)}/[x{√(1 + x²) + √(1 - x²)}]

= Lim_x→0{(1 + x²) - (1 - x²)}/[x{√(1 + x²) + √(1 - x²)}]

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

= Lim_x→0(2x/{√(1 + x²) + √(1 - x²)}

Now put x = 0, we get

= 2 × 0/(√1 + √1)

= 0

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

Solution:

We have, Lim_x→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)}}

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

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

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

Now put x = 0, we get

= 1/{2(√1 + √1)

= 1/4

Question 23. Lim_x→4{2 - √x}/(4 - x)

Solution:

We have, Lim_x→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}

= Lim_x→4{4 - x}/[(4 - x){2 + √x}]

= Lim_x→4{4 - x}/[(4 - x){2 + √x}]

= Lim_x→4(1)/{2 + √x}

Now put x = 4, we get

= 1/(2 + 2)

= 1/4

Question 24. Lim_x→a(x - a)/{√x - √a}

Solution:

We have, Lim_x→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}

= Lim_x→a[(x - a){√x - √a}]/(x - a)

= Lim_x→a{√x + √a}

Now put x = a, we get

= √a + √a

= 2√a

Question 25. Lim_x→0{√(1 + 3x) - √(1 - 3x)}/(x)

Solution:

We have, Lim_x→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)}}

= Lim_x→0{(1 + 3x) - (1 - 3x)}/[(x){√(1 + 3x) - √(1 - 3x)}]

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

= Lim_x→0(6)/{√(1 + 3x) - √(1 - 3x)}

Now put x = 0, we get

= 6/(√1 + √1)

= 6/2

= 3

Question 26. Lim_x→0{√(2 - x) - √(2 + x)}/(x)

Solution:

We have, Lim_x→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)}}

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

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

= Lim_x→0(-2)/{√(2 - x) + √(2 + x)}

Now put x = 0, we get

= (-2)/(√2 + √2)

= (-2)/(2√2)

= -1/(√2)

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

Solution:

We have, Lim_x→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)}}

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

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

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

Now put x = 1, we get

= 2/{(2)(√4 + √4)}

= 2/8

= 1/4

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

Solution:

We have, Lim_x→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

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

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

= Lim_x→1{(2x - 3)(x - 1)}/[3(x - 1)(x + 2)(√x + 1)]

= Lim_x→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. Lim_x→0{√(1 + x²) - √(1 + x)}/{√(1 + x³) - √(1 + x)}

Solution:

We have, Lim_x→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. Lim_x→1{x²- √x}/{√x - 1}

Solution:

We have, Lim_x→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

= Lim_x→1{√x(x√x -1)}/{√x - 1}

= Lim_x→1{√x(x^3/2- 1)}/{√x - 1}

= Lim_x→1[√x{(√x)³- 1}]/{√x - 1}

= Lim_x→1[(√x)(√x - 1)(x + √x + 1)]/{√x - 1}

= Lim_x→1[(√x)(x + √x + 1)]

Now put x = 1, we get

= (√1)(1 + √1 + 1)

= 3

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

Solution:

We have, Lim_h→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}}

= Lim_h→0{(x + h) - x}/[h{√(x + h) + √x}]

= Lim_h→0(h)/[h{√(x + h) + √x}]

= Lim_h→0(1)/{√(x + h) + √x}

Now put x = 0, we get

= 1/(√x + √x)

= 1/(2√x)

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

Solution:

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

= Lim_x→√10{√(7 + 2x) - √(√5 + √2)²}/{(x - √10)(x + √10)}

= Lim_x→√10{√(7 + 2x) - √(5 + 2 + 2√5√2)}/{(x - √10)(x + √10)}

= Lim_x→√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. Lim_x→√6{√(5 + 2x) - (√3 + √2)}/(x²- 6)

Solution:

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

= Lim_x→√6{√(5 + 2x) - √(√3 + √2)²}/{(x - √6)(x + √6)}

= Lim_x→√6{√(5 + 2x) - √(3 + 2 + 2√3√2)}/{(x - √6)(x + √6)}

= Lim_x→√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. Lim_x→√2{√(3 + 2x) - (√2 + 1)}/(x²- 2)

Solution:

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

= Lim_x→√2{√(3 + 2x) - √(√2 + 1)²}/{(x - √2)(x + √2)}

= Lim_x→√2{√(3 + 2x) - √(2 + 1 + 2√3)}/{(x - √2)(x + √2)}

= Lim_x→√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)

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

Evaluates the following limits:

Question 18. Lim_x→1{√(5x - 4) - √x}/(x³- 1)

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

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

Question 21. Lim_x→0{√(1 + x²) - √(1 - x²)}/(x)

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

Question 23. Lim_x→4{2 - √x}/(4 - x)

Question 24. Lim_x→a(x - a)/{√x - √a}

Question 25. Lim_x→0{√(1 + 3x) - √(1 - 3x)}/(x)

Question 26. Lim_x→0{√(2 - x) - √(2 + x)}/(x)

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

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

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

Question 30. Lim_x→1{x²- √x}/{√x - 1}

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

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

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

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

Explore

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

Evaluates the following limits:

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

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

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

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

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

Question 23. Limx→4{2 - √x}/(4 - x)

Question 24. Limx→a(x - a)/{√x - √a}

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

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

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

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

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

Question 30. Limx→1{x2 - √x}/{√x - 1}

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

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

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

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

Explore

Question 18. Lim_x→1{√(5x - 4) - √x}/(x³- 1)

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

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

Question 21. Lim_x→0{√(1 + x²) - √(1 - x²)}/(x)

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

Question 23. Lim_x→4{2 - √x}/(4 - x)

Question 24. Lim_x→a(x - a)/{√x - √a}

Question 25. Lim_x→0{√(1 + 3x) - √(1 - 3x)}/(x)

Question 26. Lim_x→0{√(2 - x) - √(2 + x)}/(x)

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

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

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

Question 30. Lim_x→1{x²- √x}/{√x - 1}

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

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

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

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