Suppose that A = {(x, y) a < x < b,c < y < d} ⊂ R2, F : A -> R . We say that F is continuous at (u, v) if the following hold :
(1) F is defined at (u, v)
(2) lim (x, y) -> (u,v) F (x, y) = L exits
(3) L = F(u, v).
Problem 1 :
Evaluate lim (x, y) -> (1, 2) g(x, y), if the limit exists, where
g(x, y) = (3x2 - xy) / (x2+y2+3)
Solution :
To check if the given limit exist
Let us approach origin along x axis.
lim (x, y) -> (x, 0) g(x, y)
= lim (x, y) -> (x, 0)(3x2 - 0) / (x2+0+3)
= 3x2 / (x2+3) ≠ ∞
Let us approach origin along y axis.
lim (x, y) -> (0, y) g(x, y)
= lim (x, y) -> (0, y)(3(0)2 - y) / (02+y+3)
= -y/(y+3) ≠ ∞
To evaluate, we have to apply x = 1 and y = 2
g(x, y) = (3x2 - xy) / (x2+y2+3)
g(x, y) -> (1, 2) = (3(1)2 - (1)(2)) / (12+22+3)
g(x, y) -> (1, 2) = (3-2) / (1+4+3)
g(x, y) -> (1, 2) = 1/8
So, the value of g(x, y) = (3x2 - xy) / (x2+y2+3) is 1/8.
Problem 2 :
Evaluate
lim (x, y) -> (0, 0) cos [(x3+y2)/(x+y+2)]
If the limit exists.
Solution :
First let us check, if the limit exists.
First let us approach (0, 0) along x-axis.
Then f(x, 0)
= cos [(x3+02)/(x+0+2)]
= cos [x3/(x+2)]
Approaching (0, 0) along y-axis.
Then f(0, y)
= cos [(03+y2)/(0+y+2)]
= cos [y2/(y+2)]
So, limit exists.
= lim (x, y) -> (0, 0) cos [(x3+y2)/(x+y+2)]
= lim (x, y) -> (0, 0) cos [(03+02)/(0+0+2)]
= lim (x, y) -> (0, 0) cos (0/2)
= lim (x, y) -> (0, 0) cos 0
= 1
Problem 3 :
Let
f(x, y) = (y2-xy)/(√x-√y)
for (x, y) ≠ 0. Show that lim (x, y) -> (0, 0) f(x, y) = 0.
Solution :
f(x, y) = (y2-xy)/(√x-√y)
By doing the direct substitution, we will get the indeterminant form.
So,
f(x, y) = lim (x, y) -> (0, 0) (y2-xy)/(√x-√y)
Multiply by the conjugate of denominator, we get
= lim (x, y) -> (0, 0) [(y2-xy)/(√x-√y)] ⋅[(√x+√y)/(√x+√y)]
= lim (x, y) -> (0, 0) [y(y-x)(√x+√y)/(x-y)]
= lim (x, y) -> (0, 0) [-y(x-y)(√x+√y)/(x-y)]
= lim (x, y) -> (0, 0) [-y(√x+√y)]
Applying x and y as 0, we get
= 0
So, the answer is 0.
Problem 4 :
Evaluate lim (x, y) -> (0, 0) cos (ex sin y/y), if the limit exists.
Solution :
= lim (x, y) -> (0, 0) cos (ex sin y/y)
To check of if the limit exists or not, we will approach the origin along x-axis.
f(x, 0)
= lim (x, y) -> (x, 0) cos (ex lim (x, y) -> (x, 0) sin y/y)
= lim (x, y) -> (x, 0) cos (ex )≠∞
we will approach the origin along y-axis.
f(0, y)
= lim (x, y) -> (0, y) cos (e0 lim (x, y) -> (0, y) sin y/y) ≠∞
So, the limit exists.
By applying x = 0 and y = 0, we get
= cos (e0 )
= cos 1
Problem 5 :
Let g(x, y) = x2y/(x4+y2) for f(x, y) ≠ 0 and f(0, 0) = 0
(i) Show that lim (x, y) -> (0, 0) g(x, y) = 0 along every line y = mx, m ∈ ℝ
(ii) Show that lim (x, y) -> (0, 0) g(x, y) = k/(1+k2) along every parabola y = kx2, m ∈ ℝ\{0}
Solution :
(i) Let us apply y = mx
g(x, y) = x2(mx)/(x4+(mx)2)
g(x, y) = mx3/(x4+m2x2)
g(x, y) = mx/(x2+m2)
lim (x, y) -> (0, 0) g(x, y) = lim (x, y) -> (0, 0) mx/(x2+m2)
= 0
(ii) Let us apply y = kx2
g(x, y) = x2(kx2)/(x4+(kx2)2)
g(x, y) = kx4/(x4+k2x4)
g(x, y) = k/(1+k2)
lim (x, y) -> (0, 0) g(x, y) = lim (x, y) -> (0, kx2) k/(1+k2)
Problem 6 :
Show that
f(x, y) = (x2-y2)/(y2+1)
is continuous at every (x, y) ∈ ℝ2.
Solution :
f(x, y) = (x2-y2)/(y2+1)
Let (x, y) -> (a, b) ∈ ℝ2.
lim (x, y) -> (0, kx2)f(x, y) = lim (x, y) -> (a,b)(x2-y2)/(y2+1)
= (a2-b2)/(b2+1)
We get the defined value.
So, the given function is continuous every (x, y).
Problem 7 :
Let g(x, y) = ey sin x/x, for x ≠ 0 and g(0, 0). Show that g is continuous at (0, 0).
Solution :
lim(x, y) -> (0, 0) g(x, y) = lim(x, y) -> (0, 0) (ey sin x/x)
= lim(x, y) -> (0, 0) (ey ) lim(x, y) -> (0, 0)(sin x/x)
= 1 (1)
= 1
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Nov 23, 24 10:01 AM
Nov 23, 24 09:45 AM
Nov 21, 24 06:13 AM