Respuestas
Respuesta dada por:
4
Lim (Sin x + Sin y)/(x+y)
x,y→(0,0)
Para resolver este límite usaremos una identidad trigonométrica:
Sea:
Sinx + Siny = Sin (x/2 + x/2) + Sin (y/2+y/2)
= 2 Sin(x/2) Cos(x/2) + 2 Sin(y/2) Cos(y/2)
=2(Sin(x/2) Cos(x/2) + Sin(y/2) Cos(y/2))
Multiplicando por 1: (Sin²x+Cos²x) y (Sin²y+Cos²y) convenientemente:
2(Sin(x/2) Cos(x/2) + Sin(y/2) Cos(y/2))
= 2(Sin(x/2) Cos(x/2) (Sin²(x/2)+Cos²(x/2)) + Sin(y/2) Cos(y/2) (Sin²(y/2)+Cos²(y/2)))
= 2 (Sin(x/2) Cos(x/2) Sin²(y/2)+Sin(x/2)Cos(x/2) Cos²(y/2) + Sin(y/2) Cos(y/2) Sin²(x/2)+ Sin(y/2) Cos(y/2) Cos²(x/2))
= 2 (Sin(x/2) Cos(x/2) Sin²(y/2)+ Sin(y/2) Cos(y/2) Sin²(x/2) + Sin(y/2) Cos(y/2) Cos²(x/2) +Sin(x/2)Cos(x/2) Cos²(y/2) )
Factorizando:
= 2[Sin(x/2)Sin(y/2) (Sin(y/2) Cos(x/2) + Sin(x/2) Cos(y/2))
+ Cos(x/2) Cos(y/2)(Sin(y/2) Cos(x/2) + Sin(x/2) Cos(y/2))]
= 2[(Sin(x/2)Sin(y/2)+Cos(x/2)Cos(y/2))(Sin(y/2)Cos(x/2)+Sin(x/2)Cos(y/2))]
= 2[Cos(x/2 - y/2) Sin(x/2 + y/2)]
Resolvamos el límite:
Lim (Sin x + Sin y)/(x+y)
x,y→(0,0)
= Lim 2[Cos(x/2 - y/2) Sin(x/2 + y/2)]/(x+y)
x,y→(0,0)
Multiplicando por 1: [(1/2)/(1/2)]
= Lim [Cos(x/2 - y/2) Sin(x/2 + y/2)]/(x/2+y/2)
x,y→(0,0)
Recordemos que:
Lim (Sin a)/a =1
x→0
Luego puesto que:
Lim x+y =1
x,y→(0,0)
Entonces:
El límite
Lim [Sin(x/2 + y/2)]/(x/2+y/2)
x,y→(0,0)
Existe y es 1.
Como el límite
Lim Cos(x/2 - y/2)
x,y→(0,0)
Existe y es 1.
Entonces podemos realizar el producto de los límites:
Lim [Cos(x/2 - y/2) Sin(x/2 + y/2)]/(x/2+y/2)
x,y→(0,0)
= Lim [Cos(x/2 - y/2) * Lim Sin(x/2 + y/2)]/(x/2+y/2)
x,y→(0,0) x,y→(0,0)
= 1*1 = 1
Luego:
Lim (Sin x + Sin y)/(x+y) = 1
x,y→(0,0)
x,y→(0,0)
Para resolver este límite usaremos una identidad trigonométrica:
Sea:
Sinx + Siny = Sin (x/2 + x/2) + Sin (y/2+y/2)
= 2 Sin(x/2) Cos(x/2) + 2 Sin(y/2) Cos(y/2)
=2(Sin(x/2) Cos(x/2) + Sin(y/2) Cos(y/2))
Multiplicando por 1: (Sin²x+Cos²x) y (Sin²y+Cos²y) convenientemente:
2(Sin(x/2) Cos(x/2) + Sin(y/2) Cos(y/2))
= 2(Sin(x/2) Cos(x/2) (Sin²(x/2)+Cos²(x/2)) + Sin(y/2) Cos(y/2) (Sin²(y/2)+Cos²(y/2)))
= 2 (Sin(x/2) Cos(x/2) Sin²(y/2)+Sin(x/2)Cos(x/2) Cos²(y/2) + Sin(y/2) Cos(y/2) Sin²(x/2)+ Sin(y/2) Cos(y/2) Cos²(x/2))
= 2 (Sin(x/2) Cos(x/2) Sin²(y/2)+ Sin(y/2) Cos(y/2) Sin²(x/2) + Sin(y/2) Cos(y/2) Cos²(x/2) +Sin(x/2)Cos(x/2) Cos²(y/2) )
Factorizando:
= 2[Sin(x/2)Sin(y/2) (Sin(y/2) Cos(x/2) + Sin(x/2) Cos(y/2))
+ Cos(x/2) Cos(y/2)(Sin(y/2) Cos(x/2) + Sin(x/2) Cos(y/2))]
= 2[(Sin(x/2)Sin(y/2)+Cos(x/2)Cos(y/2))(Sin(y/2)Cos(x/2)+Sin(x/2)Cos(y/2))]
= 2[Cos(x/2 - y/2) Sin(x/2 + y/2)]
Resolvamos el límite:
Lim (Sin x + Sin y)/(x+y)
x,y→(0,0)
= Lim 2[Cos(x/2 - y/2) Sin(x/2 + y/2)]/(x+y)
x,y→(0,0)
Multiplicando por 1: [(1/2)/(1/2)]
= Lim [Cos(x/2 - y/2) Sin(x/2 + y/2)]/(x/2+y/2)
x,y→(0,0)
Recordemos que:
Lim (Sin a)/a =1
x→0
Luego puesto que:
Lim x+y =1
x,y→(0,0)
Entonces:
El límite
Lim [Sin(x/2 + y/2)]/(x/2+y/2)
x,y→(0,0)
Existe y es 1.
Como el límite
Lim Cos(x/2 - y/2)
x,y→(0,0)
Existe y es 1.
Entonces podemos realizar el producto de los límites:
Lim [Cos(x/2 - y/2) Sin(x/2 + y/2)]/(x/2+y/2)
x,y→(0,0)
= Lim [Cos(x/2 - y/2) * Lim Sin(x/2 + y/2)]/(x/2+y/2)
x,y→(0,0) x,y→(0,0)
= 1*1 = 1
Luego:
Lim (Sin x + Sin y)/(x+y) = 1
x,y→(0,0)
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