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)

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{(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{√(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{(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{(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{(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{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[(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{(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{(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{(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

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{(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.

=

=

=

Now put x = √10, we get

=

=

=

= 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.

= 

=

=

=

Now put x = √6, we get

=

=

=

= 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.

=

=

=

=

=

Now put x = √2, we get

=

=

=

= 1/{(2√2)(√2 + 1)}

On rationalizing denominator, we get

= (√2 – 1)/{(2√2)(2 – 1)}

= (√2 – 1)/(2√2)