# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.2

**Question 1. lim **_{x → 1} (x^{2}+1)/(x+1)

_{x → 1}(x

^{2}+1)/(x+1)

**Solution: **

Using direct substitution method we get,

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lim

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

**Question 2. lim **_{x → 0} (2x^{2}+3x+4)/(x^{2}+3x+2)

_{x → 0}(2x

^{2}+3x+4)/(x

^{2}+3x+2)

**Solution:**

Using direct substitution method we get,

lim

_{x → 0}(2x^{2}+3x+4)/(x^{2}+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)**

_{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)

_{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)

_{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+x^{2})

_{x → 1 }(1+(x-1)

^{2})/(1+x

^{2})

**Solution:**

Using direct substitution method we get,

lim

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

**Question 7. lim **_{x → 0} (x^{2/3}-9)/(x-27)

_{x → 0}(x

^{2/3}-9)/(x-27)

**Solution:**

Using direct substitution method we get,

lim

_{x → 0}(x^{2/3}-9)/(x-27) = ((0)^{2/3}-9)/(0-27) = (-9)/(-27) = 9/27 =1/3

**Question 8. lim **_{x → 0} 9

_{x → 0}9

**Solution:**

Using direct substitution method we get,

lim

_{x → 0}9 =9

**Question 9. lim **_{x → 2} (3-x)

_{x → 2}(3-x)

**Solution:**

Using direct substitution method we get,

lim

_{x → 2}(3-x) = (3-2) =1

**Question 10. lim **_{x → -1} (4x^{2}+2)

_{x → -1}(4x

^{2}+2)

**Solution:**

Using direct substitution method we get,

lim

_{x → -1}(4x^{2}+2) = 4(-1)^{2}+2 = 4+2 =6

**Question 11. lim **_{x → -1} (x^{3}-3x+1)/(x-1)

_{x → -1}(x

^{3}-3x+1)/(x-1)

**Solution:**

Using direct substitution method we get,

lim

_{x → -1}(x^{3}-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)

_{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} (x^{2}-9)/(x+2)

_{x → 3}(x

^{2}-9)/(x+2)

**Solution:**

Using direct substitution method we get,

lim

_{x → 3}(x^{2}-9)/(x+2) = (32-9)/(3+2) = 0/5 =0

**Question 14. lim **_{x → 0} (ax+b)/(cx+d), d ≠** 0**

_{x → 0}(ax+b)/(cx+d), d

**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