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• RD Sharma Class 11 Solutions for Maths

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

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

Solution:

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

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

= Limx→0(x + 1)/{√(1 + x + x2) + 1}

Now put x = 0, we get

= (0 + 1)/{√(1 + 0 + 0) + 1}

= 1/2

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

Solution:

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

= Limx→0[2x{√(a + x) + √(a – x)}]/2x

= Limx→0{√(a + x) + √(a – x)}

Now put x = 0, we get

= √a + √a

= 2√a

### Question 3. Limx→0{√(a2 + x2) – a}/x2

Solution:

We have, Limx→0{√(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 = Limx→0(a2 + x2 – a2)/[x2{√(a2 + x2) + a}]

= Limx→0(x2)/[x2{√(a2 + x2) + a}]

= Limx→0(1)/{√(a2 + x2) + a}

Now put x = 0, we get

= 1/(a + a)

= 1/2a

### Question 4. Limx→0{√(1 + x) – √(1 – x)}/2x

Solution:

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

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

= Limx→0(1)/{√(1 + x) + √(1 – x)}

Now put x = 0, we get

= 1/(1 + 1)

= 1/2

### Question 5. Limx→2{√(3 – x) – 1}/(2 – x)

Solution:

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

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

= Limx→2(1)/{√(3 – x) + 1}

Now put x = 2, we get

= 1/{√(3 – 2) + 1}

= 1/(1 + 1)

= 1/2

### Question 6. Limx→3(x – 3)/{√(x – 2) – √(4 – x)}

Solution:

We have, Limx→3(x – 3)/{√(x – 2) – √(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 = Limx→3[(x – 3){√(x – 2) + √(4 – x)}]/{(x – 2) – (4 – x)}

= Limx→3[(x – 3){√(x – 2) + √(4 – x)}]/{2(x – 3)}

= Limx→3{√(x – 2) + √(4 – x)}/2

Now put x = 3, we get

= {√(3 – 2) + √(4 – 3)}/2

= (1 + 1)/2

= 1

### Question 7. Limx→0(x)/{√(1 + x) – √(1 – x)}

Solution:

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

= Limx→0[x{√(1 + x) + √(1 – x)}]/(2x)

= Limx→0{√(1 + x) + √(1 – x)}/(2)

Now put x = 0, we get

= (√1 + √1)/2

= 2/2

= 1

### Question 8. Limx→1{√(5x – 4) – √x}/(x – 1)

Solution:

We have, Limx→1{√(5x – 4) – √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{5x – 4 – x}/[(x – 1){√(5x – 4) + √x}]

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

= Limx→1(4)/{√(5x – 4) + √x}

Now put x = 1, we get

= 4/{√(5 – 4) + √1}

= 4/(1 + 1)

= 2

### Question 9. Limx→1(x – 1)/{√(x2 + 3) – 2}

Solution:

We have, Limx→1(x – 1)/{√(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 = Limx→1[(x – 1){√(x2 + 3) + 2}]/{(x2 + 3) – 4}

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

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

= Limx→1{√(x2 + 3) + 2}/{(x + 1)}

Now put x = 1, we get

= {√(1 + 3) + 2}/(1 + 1)

= (2 + 2)/2

= 2

### Question 10.  Limx→3{√(x + 3) – √6}/(x2 – 9)

Solution:

We have, Limx→3{√(x + 3) – √6}/(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 = Limx→3{(x + 3) – 6}/[(x2 – 9){√( x+ 3) + √6}]

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

= Limx→3(1)/[(x + 3){√(x + 3) + √6}]

Now put x = 3, we get  = 1/(12√6)

### Question 11. Limx→1{√(5x – 4) – √x}/(x2 – 1)

Solution:

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

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

= Limx→1(4)/[{√(5x – 4) + √x}(x + 1)]

Now put x = 1, we get

= 4/[{√(5 – 4) + √1}(1 + 1)]

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

= 4/4

= 1

### Question 12. Limx→0{√(1 + x) – 1}/x

Solution:

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

= Limx→0(x)/[x{√(1 + x) + 1}]

= Limx→0(1)/{√(1 + x) + 1}

Now put x = 0, we get

= 1/(1 + 1)

= 1/2

### Question 13. Limx→2{√(x2 + 1) – √5}/(x – 2)

Solution:

We have, Limx→2{√(x2 + 1) – √5}/(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 = Limx→2{x2 + 1 – 5}/[(x – 2){√(x2 + 1) + √5}]

= Limx→2{(x2 – 4)}/[(x – 2){√(x2 + 1) + √5}]

= Limx→2{(x – 2)(x + 2)}/[(x – 2){√(x2 + 1) + √5}]

= Limx→2{(x + 2)}/{√(x2 + 1) + √5}

Now put x = 2, we get

= 4/{√(5) + √5}

= 4/(2√5)

= 2/(√5)

### Question 14. Limx→2(x – 2)/{√x – √2}

Solution:

We have, Limx→2(x – 2)/{√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 = Limx→2[(x – 2){√x + √2}]/{x – 2}

= Limx→2{√x + √2}

Now put x = 2, we get

= √2 + √2

= 2√2

### Question 15. Limx→7{4 – √(9 + x)}/{1 – √(8 – x)}

Solution:

We have, Limx→7{4 – √(9 + x)}/{1 – √(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 + √(8 – 7)}/{4 + √(9 + 7)}

= -2/(4 + 4)

= -1/4

### Question 16. Limx→0{√(a + x) – √a}/[x{√(a2 + ax)}]

Solution:

We have,

Limx→0{√(a + x) – √a}/[x{√(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 = Limx→0{(a + x) – a}/[x{√(a2 + ax)}{√(a + x) + √a}]

= Limx→0(x)/[x{√(a2 + ax)}{√(a + x) + √a}]

= Limx→0(1)/[{√(a2 + ax)}{√(a + x) + √a}]

Now put x = 0, we get

= 1/{a.(√a + √a)}

= 1/(2a√a)

### Question 17. Limx→7(x – 5)/{√(6x – 5) – √(4x + 5)}

Solution:

We have, Limx→7(x – 5)/{√(6x – 5) – √(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 = Limx→7[{√(6x – 5) + √(4x + 5)}(x – 5)]/{(6x-5) – (4x + 5)}

= Limx→7[{√(6x – 5) + √(4x + 5)}(x – 5)]/{2(x – 5)}

= Limx→7[{√(6x – 5) + √(4x + 5)}]/(2)

Now put x = 7, we get

= {√(6 × 7 – 5) + √(4 × 7 + 5)}/(2)

= (5 + 5)/2

= 5