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# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.6 | Set 1

• Last Updated : 30 Apr, 2021

### Question 1. Limxâ†’âˆž{(3x – 1)(4x – 2)}/{(x + 8)(x – 1)}.

Solution:

We have,

Limxâ†’âˆž{(3x – 1)(4x – 2)}/{(x + 8)(x – 1)}

When x â†’ âˆž, (1/x) â†’ 0.

= (3 Ã— 4)/(1 Ã— 1)

= 12

### Question 2. Limxâ†’âˆž{(3x3 – 4x2 + 6x – 1)}/{(2x3 + x2 – 5x + 7)}.

Solution:

We have,

Limxâ†’âˆž{(3x3 – 4x2 + 6x – 1)}/{(2x3 + x2 – 5x + 7)}

=

=

When x â†’ âˆž, (1/x), (1/x2), (1/x3) â†’ 0.

= 3/2

### Question 3. Limxâ†’âˆž{(5x3 – 6)}/{âˆš(9 + 4x6)}.

Solution:

We have,

Limxâ†’âˆž{(5x3 – 6)}/{âˆš(9 + 4x6)}

=

When x â†’ âˆž, (1/x), (1/x3) â†’ 0.

= 5/âˆš4

= 5/2

### Question 4. Limxâ†’âˆž{âˆš(x2 + cx) – x}

Solution:

We have,

Limxâ†’âˆž{âˆš(x2+cx)-x}

On rationalizing numerator, we get

= Limxâ†’âˆž{(x2 + cx) – x2}/{âˆš(x2 + cx) + x}

= Limxâ†’âˆž(cx)/{âˆš(x2 + cx) + x}

= Limxâ†’âˆž(cx)/[x{âˆš(x + c/x) + 1}]

= Limxâ†’âˆž(c)/{âˆš(1 + c/x) + 1}

When x â†’ âˆž, (1/x) â†’ 0.

= c/(âˆš1 + 1)

= c/2

### Question 5. Limxâ†’âˆž{âˆš(x + 1) – âˆšx}

Solution:

We have,

Limxâ†’âˆž{âˆš(x + 1) – âˆšx}

On rationalizing numerator, we get

= Limxâ†’âˆž{(x+1)-x}/{âˆš(x+1)+âˆšx}

= Limxâ†’âˆž(1)/{âˆš(x+1)+âˆšx}

When x â†’ âˆž, (1/x) â†’ 0.

= 0

### Question 6. Limxâ†’âˆž{âˆš(x2 + 7x) – x}

Solution:

We have,

Limxâ†’âˆž{âˆš(x2 + 7x) – x}

On rationalizing numerator, we get

= Limxâ†’âˆž{(x2+7x)-x2}/{âˆš(x2+7x)+x}

= Limxâ†’âˆž(7x)/{âˆš(x2+7x)+x}

=

=

When x â†’ âˆž, (1/x) â†’ 0.

= 7/(âˆš1 + 1)

= 7/2

### Question 7.  Limxâ†’âˆž(x)/{âˆš(4x2 + 1) – 1}

Solution:

We have,

Limxâ†’âˆž(x)/{âˆš(4x2 + 1) – 1}

Rationalising denominator.

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

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

= Limxâ†’âˆž[{âˆš(4x2 + 1) + 1}]/(4x)

=

When x â†’ âˆž, (1/x2) â†’ 0.

= âˆš4/4

= 2/4

= 1/2

### Question 8. Limnâ†’âˆž(n2)/{1 + 2 + 3 + 4 + ……………. + n}

Solution:

We have,

Limnâ†’âˆž(n2)/{1 + 2 + 3 + 4 + ……………. + n}

=

= Limnâ†’âˆž(2n)/(n+1)

= Limnâ†’âˆž(2)/(1+1/n)

When n â†’ âˆž, (1/n) â†’ 0

= 2/(1 + 0)

= 2

### Question 9. Limxâ†’âˆž(3x-1 + 4x-2)/(5x-1 + 6x-2)

Solution:

We have,

Limxâ†’âˆž(3x-1 + 4x-2)/(5x-1 + 6x-2)

When x â†’ âˆž, (1/x) â†’ 0.

= 3/5

### Question 10. Limxâ†’âˆž{âˆš(x2 + a2) – âˆš(x2 + b2)}/{âˆš(x2 + c2) – âˆš(x2 + d2)}

Solution:

We have,

Limxâ†’âˆž{âˆš(x2 + a2) – âˆš(x2 + b2)}/{âˆš(x2 + c2) – âˆš(x2 + d2)}

On rationalizing numerator and denominator, we get

=

=

=

=

When x â†’ âˆž, (1/x2) â†’ 0.

=

= (a2 – b2)/(c2 – d2)

### Question 11. Limnâ†’âˆž{(n + 2)! + (n + 1)!}/{(n + 2)! – (n + 1)!}.

Solution:

We have,

Limnâ†’âˆž{(n + 2)! + (n + 1)!}/{(n + 2)! – (n + 1)!}

= Limnâ†’âˆž{(n + 2)(n + 1)! + (n + 1)!}/{(n + 2)(n + 1)! – (n + 1)!}

= Limnâ†’âˆž[(n + 1)!{(n + 2) + 1}]/[(n + 1)!{(n + 2) – 1}]

= Limnâ†’âˆž(n + 3)/(n + 1)

= Limnâ†’âˆž[n(1 + 3/n)]/[n(1 + 1/n)]

When n â†’ âˆž, (1/n) â†’ 0.

= 1/1

= 1

### Question 12. Limxâ†’âˆž[x{âˆš(x2 + 1) – âˆš(x2 – 1)}]

Solution:

We have,

Limxâ†’âˆž[x{âˆš(x2 + 1) – âˆš(x2 – 1)}]

On rationalizing numerator, we get

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

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

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

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

When x â†’ âˆž, (1/x2) â†’ 0.

= 2/(âˆš1 + âˆš1)

= 2/2

= 1

### Question 13.  Limxâ†’âˆž[âˆš(x + 2){âˆš(x + 1) – âˆšx}]

Solution:

We have,

Limxâ†’âˆž[âˆš(x + 2){âˆš(x + 1) – âˆšx}]

On rationalizing numerator, we get

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

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

= Limxâ†’âˆž[xâˆš(1 + 2/x)]/[x{âˆš(1 + 1/x) + âˆš1}]

= Limxâ†’âˆž[âˆš(1 + 2/x)]/{âˆš(1 + 1/x) + âˆš1}

When x â†’ âˆž, (1/x) â†’ 0.

= 1/(âˆš1 + âˆš1)

= 1/2

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