Negative Infinite To Positive Infinity

APPLIED MATHS
Exercise 1.
Problem 1)
You can realize that f(x,0) =f(0,y)=0 as you approach the origin. This is an evidence
that as you approach the origin, from the two axes, both cases are realized.
Limx→0f(x,0)=0 ,and limy→0f(0,y)=0
However, it is wrong to conclude that the limit is equal to zero .As you move closer
to (0,0) on the the path of the line,y=mx(with m ∈ R),you get;
limx→0f(x,mx)= lim→0
x2
a2 + ( mx )2
b2 -1
As the result depends on m,the limit does not exist.Hence,we cannot use the Dini’s
theorem at every point of solution of f(x,y)=0.
Problem 2)
Consider the function f(x,y)= e6 x2
+ 2 y2−1
3 x2 + y2
Assume that t =3x2 +y2 ,and you will get ,
e6 x2
+ 2 y2−1
3 x2 + y2 = e2 t −1
t
But ,
When f(x,y) →(0,0),it implies that also t→0.
Hence,you will have;
Lim(x,y0→0f(x,y) = Limt→0
e2 t −1
t =2
Because the result does not depend on any factor m,the limit does exist.Thus ,we
can use the Dini’s theorem at every solution of f(x,y)=0
Problem 3)

Consider =-3x +10y=25 such that -3=b,10=b and 25 =cbe an ellipse equation.
Y= 3
10 + 5
2 since a2m2 +b2 =c2
This implies that m= 3
10 and c= 5
2
But 9
100 a2 + b2 = 25
4 ,which implies that b2= 25
4 - 9
100 a2
And this implies,
9b2 + 64
25 a2 =a2b2
→9( 25
4 - 9
100 ) + 64
25 a2 = a2( 25
4 - 9
100 a2)→a2 -50a2 +625 =0
Hence a2 =25
This implies that ;
b 2 = 25
4 - 9
100 *25 =4
Hence The equation of an ellipse x2
25 + y2
4 =1
xx 0
a2 + yy 0
b2 -1=0
Exercise 2
Problem 1.
Natural domain is the maximum set of values for which a function is defined.
F(x,y)=In(x+y2)
F(0,0)=Undefined
F(-1,1)=imposible
F(1,-1)=F(1,1)=0.6931
F(0,1)=f(1,0)=0