# Engineering Mathematics Questions and Answers – Limits and Derivatives of Several Variables – 2

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Limits and Derivatives of Several Variables – 2”.

1. Two men on a surface want to meet each other. They have taken the point (0, 0) as meeting point. The surface is 3-D and its equation is f(x,y) = $$\frac{x^{\frac{-23}{4}}y^9}{x+(y)^{\frac{4}{3}}}$$. Given that they both play this game infinite number of times with their starting point as (908, 908) and (90, 180)
(choosing a different path every time they play the game). Will they always meet?
a) They will not meet every time
b) They will meet every time
c) Insufficient information
d) They meet with probability 12

Explanation: The question is asking us to simply find the limit of the given function exists as the pair (x, y) tends to (0, 0) (The two men meet along different paths taken or not)
Thus, put x = t : y = a(t)34
$$=lt_{(x,y)\rightarrow(0,0)}=lt_{t\rightarrow 0}\frac{a^9.t^{\frac{27}{4}}.t^{\frac{-23}{4}}}{t+a^{\frac{4}{3}}.t^{\frac{1}{1}}}$$
$$=lt_{t\rightarrow 0}\frac{t}{t} \times \frac{a^9}{1+a^{\frac{4}{3}}}$$
$$=lt_{t\rightarrow 0} \frac{a^9}{1+a^{\frac{4}{3}}}$$
By putting different values of a we get different limits
Thus, there are many paths that do not go to the same place.
Hence, They will not meet every time is the right answer.

2. Find $$lt_{(x,y,z)\rightarrow(0,0,0)}\frac{y^2.z^2}{x^3+x^2.(y)^{\frac{4}{3}}+x^2.(z)^{\frac{4}{3}}}$$
a) 1
b) 0
c) ∞
d) Does Not Exist

Explanation: Put x = t : y = a1 * t34 : z = a2 * t34
$$lt_{(x,y,z)\rightarrow(0,0,0)}\frac{(a_1)^2.t^{\frac{3}{2}}.(a_2)^2.t^{\frac{3}{2}}}{t^3+t^2.t.(a_1)^{\frac{4}{3}}+t^2.t.(a_2)^{\frac{4}{3}}}$$
$$lt_{(x,y,z)\rightarrow(0,0,0)}\frac{t^3}{t^3}\times \frac{(a_1)^2.(a_2)^2}{1+(a_1)^{\frac{4}{3}}+(a_2)^{\frac{4}{3}}}$$
$$lt_{(x,y,z)\rightarrow(0,0,0)} \frac{(a_1)^2.(a_2)^2}{1+(a_1)^{\frac{4}{3}}+(a_2)^{\frac{4}{3}}}$$
By varying a1 : a2 one can get different limit values.

3. Find $$lt_{(x,y,z)\rightarrow(0,0,0)}\frac{sin(x).sin(y)}{x.z}$$
a) ∞
b) 13
c) 1
d) Does Not Exist

Explanation: Put x = t : y = at : z = t
=$$lt_{t\rightarrow 0}\frac{sin(t).sin(at)}{t^2}$$
=$$lt_{t\rightarrow 0}\frac{sin(t)}{t}\times (a) \times lt_{t\rightarrow 0}\frac{sin(at)}{at}$$
= (1) * (a) * (1) = a

4. Find $$lt_{(x,y,z)\rightarrow(2,2,4)}\frac{x^2+y^2-z^2+2xy}{x+y-z}$$
a) ∞
b) 123
c) 9098
d) 8

Explanation: Simplifying the expression yields
$$lt_{(x,y,z)\rightarrow(0,0,0)}\frac{(x+y)^2-z^2}{(x+y)-z}$$
$$lt_{(x,y,z)\rightarrow(0,0,0)}\frac{(x+y+z).(x+y-z)}{(x+y-z)}$$
$$lt_{(x,y,z)\rightarrow(0,0,0)}(x+y+z)=2+2+4$$
=8

5. Find $$lt_{(x,y,z,w)\rightarrow(0,0,0,0)}\frac{x^{-6}.y^2.(z.w)^3}{x+y^2+z-w}$$
a) 1990
b) ∞
c) Does Not Exist
d) 0

Explanation: Put x = t : y = a1.t12 : z = a2.t : w = a3.t
$$lt_{(x,y,z,w)\rightarrow(0,0,0,0)}\frac{t^{-6}.t.(a_1)^2.t^6.(a_2)^3.(a_3)^3}{t+t.(a_1)^2+a_2.t-a_3.t}$$
$$lt_{(x,y,z,w)\rightarrow(0,0,0,0)}\frac{t}{t}\times \frac{(a_1)^2.(a_2)^3.(a_3)^3}{1+(a_1)^2+a_2-a_3}$$
$$lt_{(x,y,z,w)\rightarrow(0,0,0,0)} \frac{(a_1)^2.(a_2)^3.(a_3)^3}{1+(a_1)^2+a_2-a_3}$$
By changing the values of a1 : a2 : a3 we get different values of limit.
Hence, Does Not Exist is the right answer.
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6. Find $$lt_{(x,y,z,w)\rightarrow(3,1,1,11)}\frac{x^4+y^2+z^2+2x^2y+2yz+2x^2z-(w)^2}{x^2+y+z-w}$$
a) 700
b) 701
c) 699
d) 22

Explanation: Simplifying the expression we have
$$lt_{(x,y,z,w)\rightarrow(3,1,1,11)}\frac{(x^2+y+z)^2-(w)^2}{x^2+y+z-w}$$
$$lt_{(x,y,z,w)\rightarrow(3,1,1,11)}\frac{(x^2+y+z+w).(x^2+y+z-w)}{x^2+y+z-w}$$
$$lt_{(x,y,z,w)\rightarrow(3,1,1,11)}(x^2+y+z+w)$$=(32+1+1+11)
=9+1+1+11=22

7. Given that limit exists find $$lt_{(x,y,z)\rightarrow(-2,-2,-2)}\frac{sin((x+2)(y+5)(z+1))}{(x+2)(y+7)}$$
a) 1
b) 35
c) 12
d) 0

Explanation: Given that limit exists we can parameterize the curve
Put x = t : y = t : z = t
$$lt_{t\rightarrow -2}\frac{sin((t+2)(t+5)(t+1))}{(t+2)(t+7)}$$
$$lt_{t\rightarrow -2}\frac{sin((t+2)(t+5)(t+1))}{(t+2)(t+5)(t+1)}\times lt_{t\rightarrow -2}\frac{(t+5)(t+1)}{(t+7)}$$
$$(1)\times \frac{(-2+5)(-2+1)}{(-2+7)}$$
=$$\frac{(3).(1)}{(5)}=\frac{3}{5}$$

8. Given that limit exist find $$lt_{(x,y,z)\rightarrow(-9,-9,-9)}\frac{tan((x+9)(y+11)(z+7))}{(x+9)(y+10)}$$
a) 2
b) 1
c) 4
d) 3

Explanation: We can parameterize the curve by
x = y = z = t
$$lt_{t\rightarrow -9}\frac{tan((t+9)(t+11)(t+7))}{(t+9)(t+10)}$$
$$lt_{t\rightarrow -9}\frac{tan((t+9)(t+11)(t+7))}{(t+9)(t+11)(t+7)}\times lt_{t\rightarrow -9}\frac{tan((t+11)(t+7))}{t+10}$$
$$=\frac{(-9+11)(-9+7)}{(-9+10)}=\frac{(2)(2)}{(1)}$$
=4

9. Given that limit exists find $$lt_{(x,y,z)\rightarrow(-1,-1,-1)}\frac{tan((x-1)(y-2)(z-3))}{(x-1)(y-6)(z+7)}$$
a) 1
b) 12
c) 17
d) 27

Explanation: We can parameterize the curve by
x=y=z=t
$$lt_{t\rightarrow -1}\frac{tan((t-1)(t-2)(t-3))}{(t-1)(t-6)(t+7)}$$
$$lt_{t\rightarrow -1}\frac{tan((t-1)(t-2)(t-3))}{(t-1)(t-2)(t-3)}\times lt_{t\rightarrow -1}\frac{(t-2)(t-3)}{(t-6)(t+7)}$$
=$$\frac{(-1-2)(-1-3)}{(-1-6)(-1+7)}=\frac{(3)(4)}{(7)(6)}$$
=$$\frac{12}{42}=\frac{2}{7}$$

10. Given that limit exists find $$lt_{(x,y,z)\rightarrow(2,2,2)}\left (\frac{ln(1+\frac{xy-2x-y+z}{xz-2x-6z+12}+\frac{xz-5x-2z+10}{xy-7y-2x+14}}{(x-2)(y-2)(z-2)}\right )$$
a) ∞
b) 1
c) 0
d) ln(45)

Explanation: We can parameterize the curve by
x = y = z = t
$$lt_{t\rightarrow 2}\left (\frac{ln(1+\frac{xy-2x-y+z}{xz-2x-6z+12}+\frac{xz-5x-2z+10}{xy-7y-2x+14}}{(x-2)(y-2)(z-2)}\right )$$
$$=lt_{t\rightarrow 2}\left (\frac{ln(1+\frac{t}{(t-6)}+\frac{(t-5)}{(t-7)}}{(t-2)^3}\right )$$
$$=lt_{t\rightarrow 2}\left (\frac{ln(1+\frac{2}{(t-6)}+\frac{(2-5)}{(2-7)})}{(2-2)^3}\right )=\frac{ln(\frac{4}{5})}{0}\rightarrow\infty$$

11. Given that limit exists $$lt_{(x,y,z)\rightarrow(0,0,0)}\left (\frac{cos(\frac{\pi}{2}-x).tan(y).cot(\frac{\pi}{2}-z)}{sin(x).sin(y).sin(z)}\right )$$
a) 99
b) 0
c) 1
d) 100

Explanation: Put x = y = z = t
$$lt_{t\rightarrow 0}\left (\frac{cos(\frac{\pi}{2}-t).tan(y).cot(\frac{\pi}{2}-t)}{sin(x).sin(y).sin(z)}\right)$$
$$lt_{t\rightarrow 0}\frac{(sin(t))(tan^2(t))}{sin^3(t)}$$
$$=lt_{t\rightarrow 0}\frac{tan^2(t)}{sin(t)} = lt_{t\rightarrow 0}\frac{1}{cos^2(t)}$$
$$=\frac{1}{cos^2(0)}=\frac{1}{1}=1$$

12. Two men on a 3-D surface want to meet each other. The surface is given by $$f(x,y)=\frac{x^{-6}.y^7}{x+y}$$. 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 12
d) Insufficient information

Explanation: The problem asks us to find the limit of the function f(x, y) along two lines y = x and y = 2x
For the first line (first person)
x = t : y = 2t
$$=lt_{t\rightarrow 0}\frac{x^{-6}.2^7.t^7}{t+2t}=lt_{t\rightarrow 0}\frac{2^7t}{3t}$$
=$$\frac{2^7}{3}$$
For the second line (Second Person)
x = t = y
=$$lt_{t\rightarrow 0}\frac{t^{-6}.t^7}{t+t}=lt_{t\rightarrow 0}\frac{t}{2t}$$
=1/2
The limits are different and they will not meet.

13. Two men on a 3-D surface want to meet each other. The surface is given by $$f(x,y)=\frac{x^6.y^7}{x^{13}+y^{13}}$$. 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 12
d) Insufficient information

Explanation: The problem asks us to find the limit of the function f(x, y) along two lines y = x and y = 4x
For the first line (first person)
x = t : y = 4t
=$$lt_{t\rightarrow 0}\frac{t^6.4^7.t^7}{t^{13}+4^{13}t^{13}}=lt_{t\rightarrow 0}\frac{4^7t}{t(1+4^{13})}$$
=$$\frac{4^7}{1+4^{13}}$$
For the second line (Second Person)
x = t = y
=$$lt_{t\rightarrow 0}\frac{t^6.t^7}{t^{13}+t^{13}}=lt_{t\rightarrow 0}\frac{t^{13}}{2t^{13}}$$
= 12
The limits are different and the will not meet.

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) $$f(x,y) = \frac{x^7.y^8}{(x+y)}$$
b) f(x,y) = x2y7
c) $$f(x,y) = \frac{xy^2}{(x^2+y^2)}$$
d) $$f(x,y) = \frac{x^6.y^2}{(y^5+x^{10})}$$

Explanation: The curves in the given graph are parabolic and thus they can be parameterized by
x = t : y = at2
Substituting in Option $$f(x,y) = \frac{x^6.y^2}{(y^5+x^{10})}$$ we get
=$$lt_{t\rightarrow 0}\frac{t^6.a^2.t^4}{a^5.t^{10}+t^{10}}$$
$$lt_{t\rightarrow 0}\frac{t^{10}}{t^{10}}\times \frac{a^2}{a^5+1}$$
$$lt_{t\rightarrow 0} \frac{a^2}{a^5+1}$$
By varying a we get different limits

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