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

Evaluates the following limits:

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

Solution:

We have, Lim_x→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

= 

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

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

= Lim_x→0(x + 1)/{√(1 + x + x²) + 1}

Now put x = 0, we get

= (0 + 1)/{√(1 + 0 + 0) + 1}

= 1/2

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

Solution:

We have, Lim_x→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

= 

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

= Lim_x→0[2x{√(a + x) + √(a - x)}]/2x

= Lim_x→0{√(a + x) + √(a - x)}

Now put x = 0, we get

= √a + √a

= 2√a

Question 3. Lim_x→0{√(a² + x²) - a}/x²

Solution:

We have, Lim_x→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

= 

= Lim_x→0(a² + x² - a²)/[x²{√(a² + x²) + a}]

= Lim_x→0(x²)/[x²{√(a² + x²) + a}]

= Lim_x→0(1)/{√(a² + x²) + a}

Now put x = 0, we get

= 1/(a + a)

= 1/2a

Question 4. Lim_x→0{√(1 + x) - √(1 - x)}/2x

Solution:

We have, Lim_x→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

= 

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

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

= Lim_x→0(1)/{√(1 + x) + √(1 - x)}

Now put x = 0, we get

= 1/(1 + 1)

= 1/2

Question 5. Lim_x→2{√(3 - x) - 1}/(2 - x)

Solution:

We have, Lim_x→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

= 

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

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

= Lim_x→2(1)/{√(3 - x) + 1}

Now put x = 2, we get

= 1/{√(3 - 2) + 1}

= 1/(1 + 1)

= 1/2

Question 6. Lim_x→3(x - 3)/{√(x - 2) - √(4 - x)}

Solution:

We have, Lim_x→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

= 

= Lim_x→3[(x - 3){√(x - 2) + √(4 - x)}]/{(x - 2) - (4 - x)}

= Lim_x→3[(x - 3){√(x - 2) + √(4 - x)}]/{2(x - 3)}

= Lim_x→3{√(x - 2) + √(4 - x)}/2

Now put x = 3, we get

= {√(3 - 2) + √(4 - 3)}/2

= (1 + 1)/2

= 1

Question 7. Lim_x→0(x)/{√(1 + x) - √(1 - x)}

Solution:

We have, Lim_x→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

= 

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

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

= Lim_x→0{√(1 + x) + √(1 - x)}/(2)

Now put x = 0, we get

= (√1 + √1)/2

= 2/2

= 1

Question 8. Lim_x→1{√(5x - 4) - √x}/(x - 1)

Solution:

We have, Lim_x→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

= 

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

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

= Lim_x→1(4)/{√(5x - 4) + √x}

Now put x = 1, we get

= 4/{√(5 - 4) + √1}

= 4/(1 + 1)

= 2

Question 9. Lim_x→1(x - 1)/{√(x² + 3) - 2}

Solution:

We have, Lim_x→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

= 

= Lim_x→1[(x - 1){√(x² + 3) + 2}]/{(x² + 3) - 4}

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

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

= Lim_x→1{√(x² + 3) + 2}/{(x + 1)}

Now put x = 1, we get

= {√(1 + 3) + 2}/(1 + 1)

= (2 + 2)/2

= 2

Question 10.  Lim_x→3{√(x + 3) - √6}/(x² - 9)

Solution:

We have, Lim_x→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

= 

= Lim_x→3{(x + 3) - 6}/[(x² - 9){√( x+ 3) + √6}]

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

= Lim_x→3(1)/[(x + 3){√(x + 3) + √6}]

Now put x = 3, we get

= 

= 

= 1/(12√6)

Question 11. Lim_x→1{√(5x - 4) - √x}/(x² - 1)

Solution:

We have, Lim_x→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

= 

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

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

= Lim_x→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. Lim_x→0{√(1 + x) - 1}/x

Solution:

We have, Lim_x→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

= 

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

= Lim_x→0(x)/[x{√(1 + x) + 1}]

= Lim_x→0(1)/{√(1 + x) + 1}

Now put x = 0, we get

= 1/(1 + 1)

= 1/2

Question 13. Lim_x→2{√(x² + 1) - √5}/(x - 2)

Solution:

We have, Lim_x→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

= 

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

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

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

= Lim_x→2{(x + 2)}/{√(x² + 1) + √5}

Now put x = 2, we get

= 4/{√(5) + √5}

= 4/(2√5)

= 2/(√5)

Question 14. Lim_x→2(x - 2)/{√x - √2}

Solution:

We have, Lim_x→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

= 

= Lim_x→2[(x - 2){√x + √2}]/{x - 2}

= Lim_x→2{√x + √2}

Now put x = 2, we get

= √2 + √2

= 2√2

Question 15. Lim_x→7{4 - √(9 + x)}/{1 - √(8 - x)}

Solution:

We have, Lim_x→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. Lim_x→0{√(a + x) - √a}/[x{√(a² + ax)}]

Solution:

We have,

Lim_x→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

= 

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

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

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

Now put x = 0, we get

= 1/{a.(√a + √a)}

= 1/(2a√a)

Question 17. Lim_x→7(x - 5)/{√(6x - 5) - √(4x + 5)}

Solution:

We have, Lim_x→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

= 

= Lim_x→7[{√(6x - 5) + √(4x + 5)}(x - 5)]/{(6x-5) - (4x + 5)}

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

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

Now put x = 7, we get

= {√(6 × 7 - 5) + √(4 × 7 + 5)}/(2)

= (5 + 5)/2

= 5