1.=0 does not exist.

1. lim
(x , y)→0,0 ( x3− y3
3 x3 + y3 )=0 does not exist
Solution
By plugging in (x , y )→ 0,0 it becomes 0−0
0+0 =0
This therefore becomes undefined
We then evaluate the limit along different directions
For y=0
(x , y )→ x ,0
Then the limit would be lim
(x , y)→ x, 0 ( x3− y3
3 x3+ y3 )= x3−0
3 x3 +0 = x3
3 x3 =1
3
For x=0 then the limit is lim
(x , y)→0 , y ( x3 − y3
3 x3+ y3 )= 0− y3
0+ y3 =− y3
y3 =−1
The limit approaches different values along different directions
Therefore it does not exist
2)
|f ( x , y )−l|<ε
x , y →(0,0)
ε > 0 |f ( x , y )−0|<ε
|f ( x , y ) −0|<ε ⇒
| x2 y
√ x2 + y2 |<ε
| y| x2
√ x2 + y2 < ε
y2 ≤ √ x2+ y2∧0 ≤ x2 y
√ x2 + y2 ≤ 1
We have that
|y| x2
√x2 + y2 <| y|= √ y2= √x2+ y2

We see that the distance between (x,y) and (0,0) appearing on the far right
|f ( x , y )−0| we simply chooseδ to check
√ x2+ y2 <ε Thus we may
0< √ x2 + y2< δ then
√ x2+ y2 <ε holds when |f ( x , y )−0|=|y| x2
√ x2 + y2 ≤ √ x2 + y2 <ε which therefore proves that
lim
(x , y)→0,0 ( x2 y
√ x2 + y2 )=0
3)
∇ f =( ∂
∂ x + ∂
∂ y + ∂
∂ z ) f ⇒ ∂ f
∂ x + ∂ f
∂ y + ∂ f
∂ z
F(x,y,z)= √ x2+2 y2 +3 z2
f ( x , y , z )=( x2+2 y2 +3 z2)
1
2
Let u be x2+2 y2 +3 z2 using chain rule
∂ u
∂ x =2 x but f ( x , y , z ) =u
1
2 ⇒ ∂ f
∂ u = 1
2 u
−1
2 ⇒ 1
2 (x ¿¿ 2+2 y2+ 3 z2 )
−1
2 ¿
∂ f
∂ x = ∂ u
∂ x . ∂ f
∂ u =2 x × 1
2 (x¿ ¿ 2+ 2 y2 +3 z2)
−1
2 = x
√ ¿ ¿ ¿ ¿ ¿
Same case to ∂ f
∂ y = 1
2 × 1
√ ¿ ¿ ¿ ¿
∂ f
∂ z = 1
2 × 1
√ ¿ ¿ ¿ ¿
∂ f
∂ z = 3 z
√¿ ¿ ¿ ¿

∇ f = ∂ f
∂ x + ∂ f
∂ y + ∂ f
∂ z = x
√ (x¿¿ 2+2 y2 +3 z2)+ 2 y
√ (x¿ ¿2+ 2 y2 +3 z2)+ 3 z
√ ¿ ¿ ¿ ¿ ¿
¿
∇ f = x + y + z
√ ¿ ¿ ¿ ¿
∇ f at point ( 1,1,1 ) is 1+1+1
√ ¿ ¿ ¿ ¿
4)
∂ z
∂ x ∧∂ z
∂ y of z
x2+ y2−2 y+1−z2=0
z2=x2+ y2−2 y+ 1
Differentiating implicitly 2 z ∂ z
∂ x =2 x
∂ z
∂ x = x
z ⇒ x
√x2 + y2−2 y +1
2 z ∂ z
∂ y =2 y −2
Therefore ∂ z
∂ y = 2 y −2
2 z ⇒ y−1
z ⇒ y−1
√ x2+ y2 −2 y +1

ALGEBRA
5 (a)
Let multiply by a scalar factor of say 3
Then considering the vector in S undergoes a scalar multiplication
x1=( 3
0 ) ⇒ 3 ( 3
0 )= ( 9
0 )
x2=( 0
2 ) ⇒ 3 ( 0
2 )= ( 0
6 )
x3=( 0
0 ) ⇒ 3 ( 0
0 )= ( 0
0 )
This is closed under multiplication since S under scalar multiplication belongs to R2 as it in S
b)
In order to establish this ( 3
0 ) + ( 0
0 )= ( 3
0 ) this is not closed under addition because in order the first part S
to be in the second part then it needs a counterexample 3 ≠ 0 ×3+1 ?
This therefore means that it is not closed under addition
c.
S is not a subspace of R2 since it is not closed under addition
6.
PB ← ε =[ [ ε1 ] B [ ε2 ] B]
PB ← ε = [ 1 1
0 1⋮ 1 0
0 0 ]
To find the coordinates x1 , x2∈ε1 x1 b1+ x2 b2=ε1
x1 ( 1
0 ) +x2 ( 1
1 )=( 1
0 )
Use gauss Jordan method
Then PB ← ε = [ 1 0
0 1 ⋮ 1 0
0 0 ]
PB ← ε=[1 0
0 0 ]

7( a)
R3 → R2
∀ x=
( x1
x2
x3
)ϵ R3
( x1−x2
x1 + x2−x3 )
b2=
(1
3
2)
T ( b2 ) = (−2
2 ) ϵ R2
b) Given that T ( b3 ) = ( 5
1 )
Find (T ( b3 ) C
b3=
( 2
−3
2 )
( x1−x2
x1 + x2−x3 ) Given that c1= ( 1
1 ) ∧c2= ( 1
−1 )
(T ( b3 ) C=( 5
1 ) c Where c (c1= ( 1
1 )∧c2= ( 1
−1 ))
T ( b3 ) C=(1 1
1 −1 )(5
1 )=(6
4 )