第11类RD Sharma解决方案–第29章限制–练习29.2

问题1. lim _x→1 (x ² +1)/(x + 1)

解决方案：

Using direct substitution method we get,

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

问题2. lim _x→0 (2x ² + 3x + 4)/(x ² + 3x + 2)

解决方案：

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

问题3. lim _x→3 ( √ (2x + 3))/(x + 3)

解决方案：

Using direct substitution method we get,

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

问题4. lim _x→1 (√(x + 8))/(√x)

解决方案：

Using direct substitution method we get,

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

问题5. lim _x→a (((√x)+(√a))/(x + a)

解决方案：

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

问题6. lim _x→1 (1+(x-1) ² )/(1 + x ² )

解决方案：

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

问题7. lim _x→0 (x ^2/3 -9)/(x-27)

解决方案：

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

问题8. lim _x→0 9

解决方案：

Using direct substitution method we get,

lim _{x → 0}   9 = 9

问题9. lim _x→2 (3-x)

解决方案：

Using direct substitution method we get,

lim _{x → 2} (3-x) = (3-2) = 1

问题10.lim _x→-1 (4x ² +2)

解决方案：

Using direct substitution method we get,

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

问题_{11.lim x→-1} (x ³ -3x + 1)/(x-1)

解决方案：

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

问题12. lim _x→0 (3x + 1)/(x + 3)

解决方案：

Using direct substitution method we get,

lim _{x → 0} (3x+1)/(x+3) = (3(0)+1)/(0+3) = 1/3

问题13. lim _x→3 (x ² -9)/(x + 2)

解决方案：

Using direct substitution method we get,

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

问题14. lim _x→0 (ax + b)/(cx + d)，d ≠ 0

解决方案：

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