LIMIT AND CONTINUITY OF FUNCTIONS OF TWO VARIABLES

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 (e^x sin y/y), if the limit exists.

Solution :

=  lim _{(x, y) -> (0, 0)} cos (e^x 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 (e^x lim _{(x, y) -> (x, 0)} sin y/y)

=  lim _{(x, y) -> (x, 0)} cos (e^x )≠∞

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, kx}²₎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)  =  e^y 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) (e^y sin x/x)

=  lim(x, y) -> (0, 0) (e^y ) lim(x, y) -> (0, 0)(sin x/x)

=  1 (1)

=  1

Problem 8 :

Is

f (x ; y) = x²y + 3 x³ y⁴ - x + 2y

continuous at (0; 0) ? Where is it continuous?

Solution :

f (x ; y) is a polynomial function, therefore it is continuous on R². In particular, it is continuous at (0;0).

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