LIMIT AND CONTINUITY OF FUNCTIONS OF TWO VARIABLES

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 (elim (x, y) -> (x, 0) sin y/y)

=  lim (x, y) -> (x, 0) cos (e)

we will approach the origin along y-axis.

f(0, y)

=  lim (x, y) -> (0, y) cos (elim (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 ∈ 

(ii)  Show that lim (x, y) -> (0, 0)  g(x, y)  =  k/(1+k2) along every parabola y = kx2, ∈ 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, kx2k/(1+k2)

Problem 6 :

Show that

f(x, y)  =  (x2-y2)/(y2+1)

is continuous at every (x, y) ∈ 2.

Solution :

f(x, y)  =  (x2-y2)/(y2+1)

Let (x, y) -> (a, b) ∈ 2.

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

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