a) 1
b) 0
c) ∞
d) Does Not Exist

Answer: d
Explanation: Put x = t : y = a₁ * t^3⁄₄ : z = a₂ * t^3⁄₄
lt(x,y,z)→(0,0,0)(a1)2.t32.(a2)2.t32t3+t2.t.(a1)43+t2.t.(a2)43
lt(x,y,z)→(0,0,0)t3t3×(a1)2.(a2)21+(a1)43+(a2)43
lt(x,y,z)→(0,0,0)(a1)2.(a2)21+(a1)43+(a2)43
By varying a₁ : a₂ one can get different limit values.

Q 3. Find lt(x,y,z)→(0,0,0)sin(x).sin(y)x.z
a) ∞
b) ¹⁄₃
c) 1
d) Does Not Exist

Answer: d
Explanation: Put x = t : y = at : z = t
=ltt→0sin(t).sin(at)t2
=ltt→0sin(t)t×(a)×ltt→0sin(at)at
= (1) * (a) * (1) = a

a) ∞
b) 123
c) 9098
d) 8

Answer: d
Explanation: Simplifying the expression yields
lt(x,y,z)→(0,0,0)(x+y)2−z2(x+y)−z
lt(x,y,z)→(0,0,0)(x+y+z).(x+y−z)(x+y−z)
lt(x,y,z)→(0,0,0)(x+y+z)=2+2+4
=8

Q 5. Find lt(x,y,z,w)→(0,0,0,0)x−6.y2.(z.w)3x+y2+z−w
a) 1990
b) ∞
c) Does Not Exist
d) 0

Answer: c
Explanation: Put x = t : y = a₁.t^1⁄₂ : z = a₂.t : w = a₃.t
lt(x,y,z,w)→(0,0,0,0)t−6.t.(a1)2.t6.(a2)3.(a3)3t+t.(a1)2+a2.t−a3.t
lt(x,y,z,w)→(0,0,0,0)tt×(a1)2.(a2)3.(a3)31+(a1)2+a2−a3
lt(x,y,z,w)→(0,0,0,0)(a1)2.(a2)3.(a3)31+(a1)2+a2−a3
By changing the values of a₁ : a₂ : a₃ we get different values of limit.
Hence, Does Not Exist is the right answer.

a) 700
b) 701
c) 699
d) 22

Answer: d
Explanation: Simplifying the expression we have
lt(x,y,z,w)→(3,1,1,11)(x2+y+z)2−(w)2x2+y+z−w
lt(x,y,z,w)→(3,1,1,11)(x2+y+z+w).(x2+y+z−w)x2+y+z−w
lt(x,y,z,w)→(3,1,1,11)(x2+y+z+w)=(3²+1+1+11)
=9+1+1+11=22

Q 7. Given that limit exists find lt(x,y,z)→(−2,−2,−2)sin((x+2)(y+5)(z+1))(x+2)(y+7)
a) 1
b) ³⁄₅
c) ¹⁄₂
d) 0

Answer: b
Explanation: Given that limit exists we can parameterize the curve
Put x = t : y = t : z = t
ltt→−2sin((t+2)(t+5)(t+1))(t+2)(t+7)
ltt→−2sin((t+2)(t+5)(t+1))(t+2)(t+5)(t+1)×ltt→−2(t+5)(t+1)(t+7)
(1)×(−2+5)(−2+1)(−2+7)
=(3).(1)(5)=35

Q 8. Given that limit exist find lt(x,y,z)→(−9,−9,−9)tan((x+9)(y+11)(z+7))(x+9)(y+10)
a) 2
b) 1
c) 4
d) 3

Answer: c
Explanation: We can parameterize the curve by
x = y = z = t
ltt→−9tan((t+9)(t+11)(t+7))(t+9)(t+10)
ltt→−9tan((t+9)(t+11)(t+7))(t+9)(t+11)(t+7)×ltt→−9tan((t+11)(t+7))t+10
=(−9+11)(−9+7)(−9+10)=(2)(2)(1)
=4

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Most Important Engineering Mathematics Question

Q 9. Given that limit exists find lt(x,y,z)→(−1,−1,−1)tan((x−1)(y−2)(z−3))(x−1)(y−6)(z+7)
a) 1
b) ¹⁄₂
c) ¹⁄₇
d) ²⁄₇

Answer: d
Explanation: We can parameterize the curve by
x=y=z=t
ltt→−1tan((t−1)(t−2)(t−3))(t−1)(t−6)(t+7)
ltt→−1tan((t−1)(t−2)(t−3))(t−1)(t−2)(t−3)×ltt→−1(t−2)(t−3)(t−6)(t+7)
=(−1−2)(−1−3)(−1−6)(−1+7)=(3)(4)(7)(6)
=1242=27

Q 10. Given that limit exists find lt(x,y,z)→(2,2,2)(ln(1+xy−2x−y+zxz−2x−6z+12+xz−5x−2z+10xy−7y−2x+14(x−2)(y−2)(z−2))
a) ∞
b) 1
c) 0
d) ln(⁴⁄₅)

Answer: a
Explanation: We can parameterize the curve by
x = y = z = t
ltt→2(ln(1+xy−2x−y+zxz−2x−6z+12+xz−5x−2z+10xy−7y−2x+14(x−2)(y−2)(z−2))
=ltt→2(ln(1+t(t−6)+(t−5)(t−7)(t−2)3)
=ltt→2(ln(1+2(t−6)+(2−5)(2−7))(2−2)3)=ln(45)0→∞

Q 11. Given that limit exists lt(x,y,z)→(0,0,0)(cos(π2−x).tan(y).cot(π2−z)sin(x).sin(y).sin(z))
a) 99
b) 0
c) 1
d) 100
view answer

Q 12. Two men on a 3-D surface want to meet each other. The surface is given by f(x,y)=x−6.y7x+y. They make their move horizontally or vertically with the X-Y plane as their reference. It was observed that one man was initially at (200, 400) and the other at (100, 100). Their meet point is decided as (0, 0). Given that they travel in straight lines, will they meet?
a) They will meet
b) They Will not meet
c) They meet with probability ¹⁄₂
d) Insufficient information
view answer

Q 13. Two men on a 3-D surface want to meet each other. The surface is given by f(x,y)=x6.y7x13+y13. They make their move horizontally or vertically with the X-Y plane as their reference. It was observed that one man was initially at (400, 1600) and the other at (897, 897). Their meet point is decided as (0, 0). Given that they travel in straight lines, will they meet?
a) They will meet
b) They will not meet
c) They meet with probability ¹⁄₂
d) Insufficient information

Q 14. Observe the figure. It is given that the function has no limit as (x, y) → (0 ,0) along the paths given in the figure. Then which of the following could be f(x, y)
a) f(x,y)=x7.y8(x+y)
b) f(x,y) = x²y⁷
c) f(x,y)=xy2(x2+y2)
d) f(x,y)=x6.y2(y5+x10)

Answer: d
Explanation: The curves in the given graph are parabolic and thus they can be parameterized by
x = t : y = at²
Substituting in Option f(x,y)=x6.y2(y5+x10) we get
=ltt→0t6.a2.t4a5.t10+t10
ltt→0t10t10×a2a5+1
ltt→0a2a5+1
By varying a we get different limits

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