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Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.2
• Last Updated : 03 Jan, 2021

### Question 1. lim x → 1 (x2+1)/(x+1)

Solution:

Using direct substitution method we get,

lim x → 1 (x2+1)/(x+1) = (12+1)/(1+1) = 2/2 = 1

### Question 2. lim x → 0 (2x2+3x+4)/(x2+3x+2)

Solution:

Using direct substitution method we get,

lim x → 0 (2x2+3x+4)/(x2+3x+2) = (2(0)2+3(0)+4)/((0)2+3(0)+2) = 4/2 = 2

### Question 3. lim x → 3 (√(2x+3))/(x+3)

Solution:

Using direct substitution method we get,

lim x → 3 (√(2x+3))/(x+3) = (√(2(3)+3))/(3+3) = (√9)/6 = 3/6 = 1/2

### Question 4. lim x → 1 (√(x+8))/(√x)

Solution:

Using direct substitution method we get,

lim x → 1 (√(x+8))/(√x) = (√(1+8))/(√1) = (√9)/(1) = 3/1 = 3

### Question 5. lim x → a ((√x)+(√a))/(x+a)

Solution:

Using direct substitution method we get,

lim x → a ((√x)+(√a))/(x+a) = ((√a)+(√a))/(a+a) = (2√a)/(2a) = (√a)/((√a)2) = 1/√a

### Question 6. lim x → 1   (1+(x-1)2)/(1+x2)

Solution:

Using direct substitution method we get,

lim x → 1(1+(x-1)2)/(1+x2) = (1+(1-1)2)/(1+12) = (1+0)/2 = 1/2

### Question 7. lim x → 0 (x2/3-9)/(x-27)

Solution:

Using direct substitution method we get,

lim x → 0 (x2/3-9)/(x-27) = ((0)2/3-9)/(0-27) = (-9)/(-27) = 9/27 = 1/3

### Question 8. lim x → 0 9

Solution:

Using direct substitution method we get,

lim x → 0   9 = 9

### Question 9. lim x → 2 (3-x)

Solution:

Using direct substitution method we get,

lim x → 2 (3-x) = (3-2) = 1

### Question 10. lim x → -1 (4x2+2)

Solution:

Using direct substitution method we get,

lim x → -1 (4x2+2) = 4(-1)2+2 = 4+2 = 6

### Question 11. lim x → -1 (x3-3x+1)/(x-1)

Solution:

Using direct substitution method we get,

lim x → -1 (x3-3x+1)/(x-1) = ((-1)3-3(-1)+1)/(-1-1) = (-1+3+1)/(-2) = -3/2

### Question 12. lim x → 0   (3x+1)/(x+3)

Solution:

Using direct substitution method we get,

lim x → 0 (3x+1)/(x+3) = (3(0)+1)/(0+3) = 1/3

### Question 13. lim x → 3 (x2-9)/(x+2)

Solution:

Using direct substitution method we get,

lim x → 3 (x2-9)/(x+2) = (32-9)/(3+2) = 0/5 = 0

### Question 14. lim x → 0 (ax+b)/(cx+d), d ≠ 0

Solution:

Using direct substitution method we get,

lim x → 0 (ax+b)/(cx+d) = (a(0)+b)/(c(0)+d) = (0+b)/(0+d) = b/d

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