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
100a2 + b2 = 25
4 ,which implies that b2= 25
4 - 9
100a2
And this implies,
9b2 + 64
25 a2 =a2b2
→9( 25
4 - 9
100) + 64
25 a2 = a2( 25
4 - 9
100a2)→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
F(-2,-2)=0.6931
Since In(x+y2) is undefined for x=0, and also the given function In(x+y2) gives only the positive values,so
the domain of the function is (x,y)∈ (-∞,∞ ¿ U (−∞ , ∞ )
Problem 2
A is bounded between negative infinite to positive infinity. It is defined by some sets of X with real and
complex bounded values i.e R: ( x,y)∈ (-∞,∞ ¿ U (−∞ , ∞ )
Problem 3.
Lim (x,y)→0In(x+y2) =lim(x,y)→0( 1
x+ y2 ) is undefined .
You can see that f(x,y) =0 ,the derivative in (0,0) is such that
dy
dx (0,0) = dy
dx (0,0) =0
But f is differentiable at x0 if
F(x0+h) –f(x0)=∇f(x0) .h + 0(||h||) as h →0
Thus lim→0
f ( x 0+ h ) −f ( x 0 ) −∇ f ( x 0 ) . h
¿|h|∨¿ ¿
At x0=(0,0),you can say h=(h,k) since F(0,0) =0 and ∇ f ( 0,0 ) =0
Lim(h,k)→(0,0)f f ( h+ k ) −0
√ h2+ k2 =0, if A>0
Problem 4
Lim (x,y)→0In(x+y2) =lim(x,y)→0( 1
x+ y2 ) is undefined .
You can see that f(x,y) =0 ,the derivative in (0,0) is such that
dy
dx (0,0) = dy
dx (0,0) =0

But f is differentiable at x0 if
F(x0+h) –f(x0)=∇f(x0) .h + 0(||v||) as v →0
Thus lim→0
f ( x 0+ h ) −f ( x 0 ) −∇ f ( x 0 ) . h
¿|v|∨¿ ¿
At x0=(0,0),you can say h=(v,k) since F(0,0) =0 and ∇ f ( 0,0 ) =0
Lim(v,k)→(0,0)f f ( v + k ) −0
√ v +k 2 =0,
=lim(v,k)→0
1
x + v + 1
y2+ v
√1
√ ( x+ v )+( y2 +v)
=
1
0+1 + 1
0+1
√ ( 0+1 )+(0+ 1)
=
1
2
1
2
=1
Problem 5
The directional derivative is maximum in the direction of ∇f ,and the maximum value is ||∇ f ∨¿
Hence∇ f (x , y) =<fx,fy> =<Inx,Iny2>
=< 1
x , 1
y2 >
Therefore,
∇ f (1,1)=( 1
1, 1
1)=(1,1)
So that the direction is (1,1)
And the maximum value is √ 2
Problem 6
x = a + hx, y = b + hy, (x,y )∈ [1, 1].

Define f(x,y)=f(a+hx,b +hy)
F1(x,y)=fx
dx
dy +fy
dx
dy =hxf(x) +hy(y)
F11(x,y)= d f 1 ( x , y ) dy
dydx + df dy
dydx
= d
dx [hxf(x) +hyf(y)] hx + d
dy [hxf(x+hyf(y)]hy
=h2fxx+2hxhyf(x,y) +h2f(yy)
F(1,1) =f(1)(1-1) +f11(h)( 1−1
2 )
;
=f(0) +f1(1)(0) +f11(h)( 0
2 )
Problem 7
convex if for all a ∈ I, all b ∈ I, and all λ ∈ (1, 1) we have
f((1−λ)a + λb) ≤ (1 − λ)f(a) + λf(b).
At (1,1)
F(1-1)1 +1*1≤(1-1)f(1) +1(1)
F(0)1 +1 ≤0f(1) +1
1≤ 1
Hence the function is convex. It joins two x and y points such that no point lies above the graph.