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3.-Calcular si existelim

(x, y ) (3,1)F(x , y )

, donde:

F(x , y )=x

2y6xyx2+6x+9y9

(x3)4+(y1)2

Solucin:

F(x , y )=x

2y6xy+9yx2+6x9

(x3)4+(y1)2

x

(x2

6x+9 )y ( 26x+9)

(x3)4+(y1)2

F(x , y )=

F(x , y )=(x26x+9)( y1)

(x3)4+(y1)2

F(x , y )=(x3 )2(y1)2

(x3)4(y1)2

Tenemos dos caminos que pasan por (3,1):

S={(x , y )R2/y=x3 }x3

(x , y )R2 /y=1+

T=

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lim

(x , y)(3,1)S

F(x , y )=lim(x)3

F(x ,x3 )=limx 3(x3 )2(

x

11)

(x3)4+(x

31)

2

(x33 )

2

(x3 )2((x3 )

2+1

2)

limx 3

3 (x3)

9(x3)2+1

limx 3

3 (x3)

9(x3)2+1

limx 3

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x3

x3

1+( 21)

x3

1+( 21)

(x3)4+(x3 )2

x ,1+(2)=lim

x 3

F

F(x , y )=lim(x)3

lim(x , y ) (3,1)T

x3

x32

x3 4

limx 3

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x3 4

x3

4

limx 3

limx 3

(1

2)=1/2

Como:

lim

(x , y)s(3,1)

F(x , y ) lim(x, y )T (3, 1)

F(x , y)

Respuesta: lim

(x , y)(3,1)F(x , y )

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4. SiF(x , y )=

{

xy

x

2

+y2

} Si(x,!) (","). #eterminar si $ es continua en (",").

%rue&a:

$(x,!) es continua en (",")lim

(x , y)(0,0)F(x , y )=f(0,0 )=0

Calculamoslim

(x , y)(0,0)F(x , y )

lim(x , y)(0,0)

( xy

x2+y2 )

=0,para que sea continua .

Sea: S= {(x , y )R2/y=2x }

T={(x , y )R2/y=5x }