lim x-4y (x,y)- (0,0) 6y+7x .A does not exist.

Q: x²-2xy+5y²+1 lim (x,y)→(0,0) 3x²+4y²+2

A:

Q: 2x4 y lim (x,y)→(0,0) x³ + 6y²

A: We will discuss about the given limit.

Q: lim (4x – 3y)= -11 Prove that (r.,v(-2,1)"

A:

Q: x2 – 1- y lim (x,y)→(1,0) X – Vy +1

A:

Q: lim (1.)(0,0) r +y – a²y²"

A:

Q: lim (x,y)→(0,0) (x² + x²) (х, у) ->(0,0) а. 0 b. 1 с. 1/2 d. Does not exist

A:

Q: lim x+0 V1 + 3x – 1

A: SOLUTION-

Q: Find lim [x +i(2x + y)] z→1-i iz3 – 1 lim z→i z+i

A: a) Given that, limz→1-ix+i2x+y It can be written as, limx+iy→1-ix+i2x+y

Q: lim (x,y)→(0,0) In(2 + x² + y?) x² + y?

A:

Q: lim (x.y) (0,0) r + yt

A:

Q: lim (x,y)-(0,0) x²+y6

A: Evaluate: limx,y→0,0xy3x2+y6

Q: lim (x,y) (0,0) 2x3 + xy² + y x2 + 7y2

A:

Q: limy 2 X- 2

A: Here we have to evaluate limx→21x-12x-2

Q: ขน lim (x,y) → (0,0) x² +10y²

A:

Q: lim (x,y)-(2,4) x² +x/F-2y x-Vy %3D ||

A: Solve the limit

Q: lim Scy+1 1-secy a) as y→n/2 b) as y→T/4 c) as y→ 3t/2

A: a) y→π2 lim cosec y+11-sec y=cosec π2+11-sec π2=1+11-∞=-∞

Q: 22-3r 10 d) lim z5 r2 -r - 20

A:

Q: 54. lim 2 (2-x)' 55. lim y r+1 x-1 56. lim (Vx+ +2x-V-2x

A:

Q: lim (SC - S(T)) where f(x) = -x²+x+6 f(7+h)-f(7) h h→0

A: f(x)=-x2+x+6= limh→0 f(7+h)-f(7)h limh→0 -(7+h)2+(7+h)+6-[-72+7+6]h limh→0…

Q: .2 lim (x, y) (0,0) along y = x2 |3|

A: Substitute y=x2 in the function. Now the limit becomes

Q: X-4y lim (x, y)- (0,0) 6y+7x

A: limx,y→0,0x-4y6y+7x Take the path y=mx, then as x→0, y→0 limx→ox-4mx6mx+7x limx→ox1-4mx6m+7…

Q: 3x2(x – 1) (a) lim 1→∞ 5x(2x2 + 1) 2 – x3 (b) lim 2x + 5 (c) 1→-0 x4 – 2x³ + 1 2x2 (d) lim lim r--1…

A:

Q: Find the limits by rewriting the fractions first.

A: Given  lim(x,y)→(2,-4)y+4x2y-xy+4x2-4x

Q: 2x2 + 3x + 1 d. lim 1-1 22 - 2r – 3 4x + I – 3 e. lim x - 2

A:

Q: xy lim (x.y)→(4,3)x²+y²

A: find the following limit

Q: lim f(T) = 1, lim f(r) = – 00, I -2 f(-2) = %3D 4- 3- -5 -4 -3 -2 A1 2 3 4 -2 -3- -4-

A:

Q: secy-1 Lim seczy-1 a) as y → 0 b) as y – → 1/2

A:

Q: secy-1 Lim sec2y-1 a) as y → 0 b) as y → T/2

A: To determine the value of the limit limy→0secy-1sec2y-1, apply L'Hopital's rule.…

Q: x²-2xy lim (x, y)-(2,1) x2-4y2 2 3 .A O 1.B O 1 4 .C O 2 .DO

A: lim(x,y)→2,1x2-2xyx2-4y2

Q: xy lim (2,y)→(0,0) x³ + y²

A: Solution is given below..

Q: 3x²+2x+1 Q1: Using the L Hopital Rule Evaluate: lim x→0o 5x2+3x+3

A: Given: limx→∞3x2+2x+15x2+3x+3  ……1

Q: lim (x,y)-(0,0) x-y+2√x-2√√y √x - √y

A:

Q: x² + y? lim (x,y)→(0,0) /x² =? + y² + 1 – 1 A. -1 В. 2

A: In this question we will use basic of calculus i.e concept of limits of two variables.  To find…

Q: lim 人う 又+ 2x t3 sove w/o |hoptal's neue:

A: Infinity limits

Q: x - y 1. lim (x,y)¬(3,3) 1 2. lim x-0 xVx + 1

A: (1)  lim(x,y)→(3,3)x-yx-y=lim(x,y)→(3,3)x-yx-y×x+yx+y (We rationalize the denominator)…

Q: lim (x,y)→(0,0) (3 + (x² + y²) In(x² + y´

A:

Q: x*-16y* lim 14. (r,y)→ (0,0) x2 +4y2 xy 15. lim (x,y) (0,0) 3x² +2y2

A:

Q: iz° + 1) lim z→-i\ z² 2 +1,

A:

Q: lim vn² +1-n

A: Given limx→∞n2+1-1

Q: lim (x, y) → (4, -2) x2y + xy2 x? – y2

A: The given expression of the limit is: lim(x,y)→(4,-2)x2y+xy2x2-y2

Q: lim (z,y)→(0,0) (2+(x² +y²) In(x² + y²))

A: We have to find the limit of: lim(x,y)→(0,0) 2+(x2+y2)ln(x2+y2)

Q: lim 2V3. +onx 2cosx+1 %3D 6.

A: Our Aim is to evaluate limx→π623×tanx2 cosx+1 -(i)

Q: (x-a) 2. lim x→a Vx-³Va

A:

Q: x²y²z² lim (x,y,z)→(0,0,0) x² + y2 + z²

A:

Q: 2xy lim (x.y)-(0,0) x² + 2y6

A:

Q: lim (x,y)(1,2) 2x +x*y-xy 4x2-y?

A: Given,          limx,y→1,22x3+x2y-xy24x2-y2

Q: x2-2xy lim (x,y) - (2,1) x²-4y2 .A .B 1 .C 1 .D 1 4

A: We have to evaluate the following limit. lim(x,y)→(2,1)x2-2xyx2-4y2

Limits of Multivariable Functions

Multivariable calculus is an extension of single variable calculus. It involves several variables instead of just one. Assume there is an open set containing points (x0, y0), let f be a function defined in that open interval except for the points (x0, y0). The multivariable limit of this function as it approaches the points (x0, y0) and is denoted as: