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
view answer

Answer: a
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
view answer

Answer: d
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
view answer

Answer: d
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
advertisement
advertisement

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
view answer

Answer: d
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
view answer

Answer: c
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.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

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
view answer

Answer: d
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
view answer

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

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
view answer

Answer: c
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
view answer

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

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)
view answer

Answer: a
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
view answer

Answer: c
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
view answer

Answer: b
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
view answer

Answer: b
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) \(f(x,y) = \frac{x^6.y^2}{(y^5+x^{10})}\)
view answer

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 \(f(x,y) = \frac{x^6.y^2}{(y^5+x^{10})}\) 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

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

To practice all areas of Engineering Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.