# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.4 | Set 1

Last Updated : 28 Apr, 2021

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

Solution:

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

Find the limit of the given equation when x =>0.

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 – 1)/{xâˆš(1 + x + x2) + 1}

= Limxâ†’0{x(x + 1)}/{xâˆš(1 + x + x2) + 1}

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

Now put x = 0, we get

= (0 + 1)/{âˆš(1 + 0 + 0) + 1}

= 1/2

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

Solution:

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

Find the limit of the given equation when x =>0.

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

So, on rationalizing the given equation we get

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

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

= Limxâ†’0{âˆš(a + x) + âˆš(a – x)}

Now put x = 0, we get

= âˆša + âˆša

= 2âˆša

### Question 3. Limxâ†’0{âˆš(a2 + x2) – a}/x2

Solution:

We have, Limxâ†’0{âˆš(a2 + x2) – a}/x2

Find the limit of the given equation when x =>0.

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

So, on rationalizing the given equation we get

= Limxâ†’0(a2 + x2 – a2)/[x2{âˆš(a2 + x2) + a}]

= Limxâ†’0(x2)/[x2{âˆš(a2 + x2) + a}]

= Limxâ†’0(1)/{âˆš(a2 + x2) + a}

Now put x = 0, we get

= 1/(a + a)

= 1/2a

### Question 4. Limxâ†’0{âˆš(1 + x) – âˆš(1 – x)}/2x

Solution:

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

Find the limit of the given equation when x =>0.

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 – 1 + x)/[2x{âˆš(1 + x) + âˆš(1 – x)}]

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

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

Now put x = 0, we get

= 1/(1 + 1)

= 1/2

### Question 5. Limxâ†’2{âˆš(3 – x) – 1}/(2 – x)

Solution:

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

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{(3 – x) – 1}/[(2 – x){âˆš(3 – x) + 1}]

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

= Limxâ†’2(1)/{âˆš(3 – x) + 1}

Now put x = 2, we get

= 1/{âˆš(3 – 2) + 1}

= 1/(1 + 1)

= 1/2

### Question 6. Limxâ†’3(x – 3)/{âˆš(x – 2) – âˆš(4 – x)}

Solution:

We have, Limxâ†’3(x – 3)/{âˆš(x – 2) – âˆš(4 – x)}

Find the limit of the given equation

When we put x = 3, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’3[(x – 3){âˆš(x – 2) + âˆš(4 – x)}]/{(x – 2) – (4 – x)}

= Limxâ†’3[(x – 3){âˆš(x – 2) + âˆš(4 – x)}]/{2(x – 3)}

= Limxâ†’3{âˆš(x – 2) + âˆš(4 – x)}/2

Now put x = 3, we get

= {âˆš(3 – 2) + âˆš(4 – 3)}/2

= (1 + 1)/2

= 1

### Question 7. Limxâ†’0(x)/{âˆš(1 + x) – âˆš(1 – x)}

Solution:

We have, Limxâ†’0(x)/{âˆš(1 + x) – 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

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

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

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

Now put x = 0, we get

= (âˆš1 + âˆš1)/2

= 2/2

= 1

### Question 8. Limxâ†’1{âˆš(5x – 4) – âˆšx}/(x – 1)

Solution:

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

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

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

= Limxâ†’1(4)/{âˆš(5x – 4) + âˆšx}

Now put x = 1, we get

= 4/{âˆš(5 – 4) + âˆš1}

= 4/(1 + 1)

= 2

### Question 9. Limxâ†’1(x – 1)/{âˆš(x2 + 3) – 2}

Solution:

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

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[(x – 1){âˆš(x2 + 3) + 2}]/{(x2 + 3) – 4}

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

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

= Limxâ†’1{âˆš(x2 + 3) + 2}/{(x + 1)}

Now put x = 1, we get

= {âˆš(1 + 3) + 2}/(1 + 1)

= (2 + 2)/2

= 2

### Question 10.  Limxâ†’3{âˆš(x + 3) – âˆš6}/(x2 – 9)

Solution:

We have, Limxâ†’3{âˆš(x + 3) – âˆš6}/(x2 – 9)

Find the limit of the given equation

When we put x = 3, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’3{(x + 3) – 6}/[(x2 – 9){âˆš( x+ 3) + âˆš6}]

= Limxâ†’3(x – 3)/[(x – 3)(x + 3){âˆš(x + 3) + âˆš6}]

= Limxâ†’3(1)/[(x + 3){âˆš(x + 3) + âˆš6}]

Now put x = 3, we get

= 1/(12âˆš6)

### Question 11. Limxâ†’1{âˆš(5x – 4) – âˆšx}/(x2 – 1)

Solution:

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

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

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

Now put x = 1, we get

= 4/[{âˆš(5 – 4) + âˆš1}(1 + 1)]

= 4/[(1 + 1)(1 + 1)]

= 4/4

= 1

### Question 12. Limxâ†’0{âˆš(1 + x) – 1}/x

Solution:

We have, Limxâ†’0{âˆš(1 + x) – 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

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

= Limxâ†’0(x)/[x{âˆš(1 + x) + 1}]

= Limxâ†’0(1)/{âˆš(1 + x) + 1}

Now put x = 0, we get

= 1/(1 + 1)

= 1/2

### Question 13. Limxâ†’2{âˆš(x2 + 1) – âˆš5}/(x – 2)

Solution:

We have, Limxâ†’2{âˆš(x2 + 1) – âˆš5}/(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{x2 + 1 – 5}/[(x – 2){âˆš(x2 + 1) + âˆš5}]

= Limxâ†’2{(x2 – 4)}/[(x – 2){âˆš(x2 + 1) + âˆš5}]

= Limxâ†’2{(x – 2)(x + 2)}/[(x – 2){âˆš(x2 + 1) + âˆš5}]

= Limxâ†’2{(x + 2)}/{âˆš(x2 + 1) + âˆš5}

Now put x = 2, we get

= 4/{âˆš(5) + âˆš5}

= 4/(2âˆš5)

= 2/(âˆš5)

### Question 14. Limxâ†’2(x – 2)/{âˆšx – âˆš2}

Solution:

We have, Limxâ†’2(x – 2)/{âˆšx – âˆš2}

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[(x – 2){âˆšx + âˆš2}]/{x – 2}

= Limxâ†’2{âˆšx + âˆš2}

Now put x = 2, we get

= âˆš2 + âˆš2

= 2âˆš2

### Question 15. Limxâ†’7{4 – âˆš(9 + x)}/{1 – âˆš(8 – x)}

Solution:

We have, Limxâ†’7{4 – âˆš(9 + x)}/{1 – âˆš(8 – x)}

Find the limit of the given equation

When we put x = 7, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

=

=

=

Now put x = 7, we get

= -{1 + âˆš(8 – 7)}/{4 + âˆš(9 + 7)}

= -2/(4 + 4)

= -1/4

### Question 16. Limxâ†’0{âˆš(a + x) – âˆša}/[x{âˆš(a2 + ax)}]

Solution:

We have,

Limxâ†’0{âˆš(a + x) – âˆša}/[x{âˆš(a2 + ax)}]

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{(a + x) – a}/[x{âˆš(a2 + ax)}{âˆš(a + x) + âˆša}]

= Limxâ†’0(x)/[x{âˆš(a2 + ax)}{âˆš(a + x) + âˆša}]

= Limxâ†’0(1)/[{âˆš(a2 + ax)}{âˆš(a + x) + âˆša}]

Now put x = 0, we get

= 1/{a.(âˆša + âˆša)}

= 1/(2aâˆša)

### Question 17. Limxâ†’7(x – 5)/{âˆš(6x – 5) – âˆš(4x + 5)}

Solution:

We have, Limxâ†’7(x – 5)/{âˆš(6x – 5) – âˆš(4x + 5)}

Find the limit of the given equation

When we put x = 7, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Limxâ†’7[{âˆš(6x – 5) + âˆš(4x + 5)}(x – 5)]/{(6x-5) – (4x + 5)}

= Limxâ†’7[{âˆš(6x – 5) + âˆš(4x + 5)}(x – 5)]/{2(x – 5)}

= Limxâ†’7[{âˆš(6x – 5) + âˆš(4x + 5)}]/(2)

Now put x = 7, we get

= {âˆš(6 Ã— 7 – 5) + âˆš(4 Ã— 7 + 5)}/(2)

= (5 + 5)/2

= 5

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