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

### Evaluates the following limits:

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

**Solution:**

We have, Lim

_{x→0}{√(1 + x + x^{2}) – 1}/xAttention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

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free classeswhich will definitely help them in making a wise career choice in the future.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 + x^{2 }– 1)/{x√(1 + x + x^{2}) + 1}= Lim

_{x→0}{x(x + 1)}/{x√(1 + x + x^{2}) + 1}= Lim

_{x→0}(x + 1)/{√(1 + x + x^{2}) + 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}{√(a^{2 }+ x^{2}) – a}/x^{2}

**Solution:**

We have, Lim

_{x→0}{√(a^{2 }+ x^{2}) – a}/x^{2}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}(a^{2 }+ x^{2 }– a^{2})/[x^{2}{√(a^{2 }+ x^{2}) + a}]= Lim

_{x→0}(x^{2})/[x^{2}{√(a^{2 }+ x^{2}) + a}]= Lim

_{x→0}(1)/{√(a^{2 }+ x^{2}) + 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)}/2xFind 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)}/2Now 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)/{√(x^{2 }+ 3) – 2}

**Solution:**

We have, Lim

_{x→1}(x – 1)/{√(x^{2 }+ 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){√(x^{2 }+ 3) + 2}]/{(x^{2 }+ 3) – 4}= Lim

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

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

_{x→1}{√(x^{2 }+ 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}/(x^{2 }– 9)

**Solution:**

We have, Lim

_{x→3}{√(x + 3) – √6}/(x^{2 }– 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}/[(x^{2 }– 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}/(x^{2 }– 1)

**Solution:**

We have, Lim

_{x→1}{√(5x – 4) – √x}/(x^{2 }– 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^{2 }– 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}/xFind 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}{√(x^{2 }+ 1) – √5}/(x – 2)

**Solution:**

We have, Lim

_{x→2}{√(x^{2 }+ 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}{x^{2 }+ 1 – 5}/[(x – 2){√(x^{2 }+ 1) + √5}]= Lim

_{x→2}{(x^{2 }– 4)}/[(x – 2){√(x^{2 }+ 1) + √5}]= Lim

_{x→2}{(x – 2)(x + 2)}/[(x – 2){√(x^{2 }+ 1) + √5}]= Lim

_{x→2}{(x + 2)}/{√(x^{2 }+ 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{√(a^{2 }+ ax)}]

**Solution:**

We have,

Lim

_{x→0}{√(a + x) – √a}/[x{√(a^{2 }+ 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{√(a^{2 }+ ax)}{√(a + x) + √a}]= Lim

_{x→0}(x)/[x{√(a^{2 }+ ax)}{√(a + x) + √a}]= Lim

_{x→0}(1)/[{√(a^{2 }+ 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