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

Last Updated : 28 Apr, 2021

Evaluates the following limits:

Question 18. Limxโ†’1{โˆš(5x - 4) - โˆšx}/(x3 - 1)

Solution:

We have, Limxโ†’1{โˆš(5x - 4) - โˆšx}/(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_{xโ†’1}\frac{\sqrt{(5x - 4)} - \sqrt{x}}{(x^3 - 1)} \times \frac{\sqrt{(5x - 4)} + \sqrt{x}}{\sqrt{(5x - 4)} + \sqrt{x}}

= Limxโ†’1{(5x - 4) - x}/[{โˆš(5x - 4) + โˆšx}(x3 - 1)]

= Limxโ†’1{4(x - 1)}/[{โˆš(5x - 4) + โˆšx}(x-1)(x2 + x + 1)]

= Limxโ†’1(4)/[{โˆš(5x - 4) + โˆšx}(x2 + x + 1)]

Now put x = 1, we get

= 4/{(3)(โˆš1 + โˆš1)}

= 4/6

= 2/3

Question 19. Limxโ†’2{โˆš(1 + 4x) - โˆš(5 + 2x)}/(x - 2)

Solution:

We have, Limxโ†’2{โˆš(1 + 4x) - โˆš(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_{xโ†’2}\frac{\sqrt{(1 + 4x)} - \sqrt{(5 + 2x)}}{(x - 2)} \times \frac{\sqrt{(1 + 4x)} + \sqrt{(5 + 2x)}}{\sqrt{(1 + 4x)} + \sqrt{(5 + 2x)}}

= Limxโ†’2{โˆš(1 + 4x) - โˆš(5+2x)}/[(x - 2){โˆš(1 + 4x) + โˆš(5 + 2x)}]

= Limxโ†’2{(1 + 4x) - (5 + 2x)}/[(x - 2){โˆš(1 + 4x) + โˆš(5 + 2x)}]

= Limxโ†’2{2(x - 2)}/[(x - 2){โˆš(1 + 4x) + โˆš(5 + 2x)}]

= Limxโ†’2(2)/{โˆš(1 + 4x) + โˆš(5 + 2x)}

Now put x = 2, we get

= 2/{โˆš(1 + 8) + โˆš(5 + 4)}

= 2/(3 + 3)

= 1/3

Question 20. Limxโ†’1{โˆš(3 + x) - โˆš(5 - x)}/(x2 - 1)

Solution:

We have, Limxโ†’1{โˆš(3 + x) - โˆš(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_{xโ†’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)}}

= Limxโ†’1{(3 + x) - (5 - x)}/[(x2 - 1){โˆš(3 + x) + โˆš(5 - x)}]

= Limxโ†’1{2(x - 1)}/[(x - 1)(x + 1){โˆš(3 + x) + โˆš(5 - x)}]

= Limxโ†’1(2)/[(x + 1){โˆš(3 + x) + โˆš(5 - x)}]

Now put x = 1, we get

= 2/{2(2 + 2)}

= 1/4

Question 21. Limxโ†’0{โˆš(1 + x2) - โˆš(1 - x2)}/(x)

Solution:

We have, Limxโ†’0{โˆš(1 + x2) - โˆš(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_{xโ†’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)}}

= Limxโ†’0{(1 + x2) - (1 - x2)}/[x{โˆš(1 + x2) + โˆš(1 - x2)}]

= Limxโ†’0{(1 + x2) - (1 - x2)}/[x{โˆš(1 + x2) + โˆš(1 - x2)}]

= Limxโ†’0(2x2/[x{โˆš(1 + x2) + โˆš(1 - x2)}]

= Limxโ†’0(2x/{โˆš(1 + x2) + โˆš(1 - x2)}

Now put x = 0, we get

= 2 ร— 0/(โˆš1 + โˆš1)

= 0

Question 22. Limxโ†’0{โˆš(1 + x + x2) - โˆš(x + 1)}/(2x2)

Solution:

We have, Limxโ†’0{โˆš(1 + x + x2) - โˆš(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_{xโ†’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)}}

= Limxโ†’0{(1 + x + x2) - (x + 1)}/[2x2{โˆš(1 + x + x2) - โˆš(x + 1)}]

= Limxโ†’0(x2)/[2x2{โˆš(1 + x + x2) - โˆš(x + 1)}]

= Limxโ†’0(1)/[2{โˆš(1 + x + x2) - โˆš(x + 1)}]

Now put x = 0, we get

= 1/{2(โˆš1 + โˆš1)

= 1/4

Question 23. Limxโ†’4{2 - โˆšx}/(4 - x)

Solution:

We have, Limxโ†’4{2 - โˆšx}/(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_{xโ†’4}\frac{2 - โˆšx}{(4 - x)} \times \frac{2 + โˆšx}{2 + โˆšx}

= Limxโ†’4{4 - x}/[(4 - x){2 + โˆšx}]

= Limxโ†’4{4 - x}/[(4 - x){2 + โˆšx}]

= Limxโ†’4(1)/{2 + โˆšx}

Now put x = 4, we get

= 1/(2 + 2)

= 1/4

Question 24. Limxโ†’a(x - a)/{โˆšx - โˆša}

Solution:

We have, Limxโ†’a(x - a)/{โˆšx - โˆša}

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_{xโ†’a}\frac{(x - a)}{โˆšx - โˆša} \times  \frac{โˆšx + โˆša}{โˆšx + โˆša}

= Limxโ†’a[(x - a){โˆšx - โˆša}]/(x - a)

= Limxโ†’a{โˆšx + โˆša}

Now put x = a, we get

= โˆša + โˆša

= 2โˆša

Question 25. Limxโ†’0{โˆš(1 + 3x) - โˆš(1 - 3x)}/(x)

Solution:

We have, Limxโ†’0{โˆš(1 + 3x) - โˆš(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_{xโ†’0}\frac{\sqrt{(1 + 3x)} - \sqrt{(1 - 3x)}}{x} \times \frac{\sqrt{(1 + 3x)} + \sqrt{(1 - 3x)}}{\sqrt{(1 + 3x)} + \sqrt{(1 - 3x)}}

= Limxโ†’0{(1 + 3x) - (1 - 3x)}/[(x){โˆš(1 + 3x) - โˆš(1 - 3x)}]

= Limxโ†’0(6x)/[(x){โˆš(1 + 3x) - โˆš(1 - 3x)}]

= Limxโ†’0(6)/{โˆš(1 + 3x) - โˆš(1 - 3x)}

Now put x = 0, we get

= 6/(โˆš1 + โˆš1)

= 6/2

= 3

Question 26. Limxโ†’0{โˆš(2 - x) - โˆš(2 + x)}/(x)

Solution:

We have, Limxโ†’0{โˆš(2 - x) - โˆš(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_{xโ†’0}\frac{\sqrt{(2 - x)} - \sqrt{(2 + x)}}{x} \times \frac{\sqrt{(2 - x)} + \sqrt{(2 + x)}}{\sqrt{(2 - x)} + \sqrt{(2 + x)}}

= Limxโ†’0{(2 - x) - (2 + x)}/[x{โˆš(2 - x) + โˆš(2 + x)}]

= Limxโ†’0(-2x)/[x{โˆš(2 - x) + โˆš(2 + x)}]

= Limxโ†’0(-2)/{โˆš(2 - x) + โˆš(2 + x)}

Now put x = 0, we get

= (-2)/(โˆš2 + โˆš2)

= (-2)/(2โˆš2)

= -1/(โˆš2)

Question 27. Limxโ†’1{โˆš(3 + x) - โˆš(5 - x)}/(x2 - 1)

Solution:

We have, Limxโ†’1{โˆš(3 + x) - โˆš(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_{xโ†’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)}}

= Limxโ†’1{(3 + x) - (5 - x)}/[(x2 - 1){โˆš(3 + x) + โˆš(5 - x)}]

= Limxโ†’1{2(x - 1)}/[(x - 1)(x + 1){โˆš(3 + x) + โˆš(5 - x)}]

= Limxโ†’1(2)/[(x + 1){โˆš(3 + x) + โˆš(5 - x)}]

Now put x = 1, we get

= 2/{(2)(โˆš4 + โˆš4)}

= 2/8

= 1/4

Question 28. Limxโ†’1{(2x - 3)(โˆšx - 1)}/(3x2 + 3x - 6)

Solution:

We have, Limxโ†’1{(2x - 3)(โˆšx - 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

= Limxโ†’1{(2x - 3)(x - 1)}/[(3x2 + 3x - 6)(โˆšx + 1)]

= Limxโ†’1{(2x - 3)(x - 1)}/[3(x2 + x - 2)(โˆšx + 1)]

= Limxโ†’1{(2x - 3)(x - 1)}/[3(x - 1)(x + 2)(โˆšx + 1)]

= Limxโ†’1(2x - 3)/[3(x + 2)(โˆšx + 1)]

Now put x = 1, we get

= (2 - 3)/{3(3)(โˆš1 + 1)

= -1/(3 ร— 3 ร— 2)

= -1/18

Question 29. Limxโ†’0{โˆš(1 + x2) - โˆš(1 + x)}/{โˆš(1 + x3) - โˆš(1 + x)}

Solution:

We have, Limxโ†’0{โˆš(1 + x2) - โˆš(1 + x)}/{โˆš(1 + x3) - โˆš(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)}ร—\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)}ร—\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)}ร—\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}

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

Now put x = 0, we get

= (โˆš1 + โˆš1)/{1(โˆš1 + โˆš1)}

= 2/2

= 1

Question 30. Limxโ†’1{x2 - โˆšx}/{โˆšx - 1}

Solution:

We have, Limxโ†’1{x2 - โˆšx}/{โˆšx - 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

= Limxโ†’1{โˆšx(xโˆšx -1)}/{โˆšx - 1}

= Limxโ†’1{โˆšx(x3/2 - 1)}/{โˆšx - 1}

= Limxโ†’1[โˆšx{(โˆšx)3 - 1}]/{โˆšx - 1}

= Limxโ†’1[(โˆšx)(โˆšx - 1)(x + โˆšx + 1)]/{โˆšx - 1}

= Limxโ†’1[(โˆšx)(x + โˆšx + 1)]

Now put x = 1, we get

= (โˆš1)(1 + โˆš1 + 1)

= 3

Question 31. Limhโ†’0{โˆš(x + h) - โˆšx}/(h), x โ‰  0   

Solution:

We have, Limhโ†’0{โˆš(x + h) - โˆšx}/(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_{hโ†’0}\frac{\sqrt{(x + h)} - \sqrt{x}}{h} \times \frac{\sqrt{(x + h)} + \sqrt{x}}{\sqrt{(x + h)} + \sqrt{x}}

= Limhโ†’0{(x + h) - x}/[h{โˆš(x + h) + โˆšx}]

= Limhโ†’0(h)/[h{โˆš(x + h) + โˆšx}]

= Limhโ†’0(1)/{โˆš(x + h) + โˆšx}

Now put x = 0, we get

= 1/(โˆšx + โˆšx)

= 1/(2โˆšx)

Question 32. Limxโ†’โˆš10{โˆš(7 + 2x) - (โˆš5 + โˆš2)}/(x2 - 10)

Solution:

We have, Limxโ†’โˆš10{โˆš(7 + 2x) - (โˆš5 + โˆš2)}/(x2 - 10)

= Limxโ†’โˆš10{โˆš(7 + 2x) - โˆš(โˆš5 + โˆš2)2}/{(x - โˆš10)(x + โˆš10)}

= Limxโ†’โˆš10{โˆš(7 + 2x) - โˆš(5 + 2 + 2โˆš5โˆš2)}/{(x - โˆš10)(x + โˆš10)}

= Limxโ†’โˆš10{โˆš(7 + 2x) - โˆš(7 + 2โˆš10)}/{(x - โˆš10)(x + โˆš10)}

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 = โˆš10, 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/{(2โˆš10)(โˆš5 + โˆš2)}

On rationalizing denominator.

= (โˆš5 - โˆš2)/{(2โˆš10)(5 - 2)}

= (โˆš5 - โˆš2)/(6โˆš10)

Question 33. Limxโ†’โˆš6{โˆš(5 + 2x) - (โˆš3 + โˆš2)}/(x2 - 6)

Solution:

We have, Limxโ†’โˆš6{โˆš(5 + 2x) - (โˆš3 + โˆš2)}/(x2 - 6)

= Limxโ†’โˆš6{โˆš(5 + 2x) - โˆš(โˆš3 + โˆš2)2}/{(x - โˆš6)(x + โˆš6)}

= Limxโ†’โˆš6{โˆš(5 + 2x) - โˆš(3 + 2 + 2โˆš3โˆš2)}/{(x - โˆš6)(x + โˆš6)}

= Limxโ†’โˆš6{โˆš(5 + 2x) - โˆš(5 + 2โˆš6)}/{(x  -โˆš6)(x + โˆš6)}

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 = โˆš6, 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/{(2โˆš6)(โˆš3 + โˆš2)}

On rationalizing denominator, we get

= (โˆš3 - โˆš2)/{(2โˆš6)(3 - 2)}

= (โˆš3 - โˆš2)/(2โˆš6)

Question 34. Limxโ†’โˆš2{โˆš(3 + 2x) - (โˆš2 + 1)}/(x2 - 2)

Solution:

We have,  Limxโ†’โˆš2{โˆš(3 + 2x) - (โˆš2 + 1)}/(x2 - 2)

= Limxโ†’โˆš2{โˆš(3 + 2x) - โˆš(โˆš2 + 1)2}/{(x - โˆš2)(x + โˆš2)}

= Limxโ†’โˆš2{โˆš(3 + 2x) - โˆš(2 + 1 + 2โˆš3)}/{(x - โˆš2)(x + โˆš2)}

= Limxโ†’โˆš2{โˆš(3 + 2x) - โˆš(3 + 2โˆš3)}/{(x - โˆš2)(x + โˆš2)}

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 = โˆš2, 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/{(2โˆš2)(โˆš2 + 1)}

On rationalizing denominator, we get

= (โˆš2 - 1)/{(2โˆš2)(2 - 1)}

= (โˆš2 - 1)/(2โˆš2)

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