Respuestas
El valor de los limites indeterminados con radicales es :
1. Lim (x+4)/√x+4 = 0
x→ -4
2. Lim (√x-√a)/(x-a) = √a/2a
x→ a
3. Lim x+1 /√6x²+3 +3x = 1
x→-1
4. Lim (√x+2 - 2) /(x-2) = 1/4
x→2
5. Lim( √x²+25 - 5) /(√x² +16 -4 ) = 4/5
x→0
Los valores de los limites indeterminados con radicales se calculan mediante la aplicación de la conjugada, de la siguiente manera :
1. Lim (x+4)/√x+4 = (-4+4)/√(-4+4) = 0/0 ind
x→ -4
Lim (x+4)/√x+4 *( √x+4)/( √x+4) = Lim (x+4) * √x+4/(x+4) =
x→ -4 x→ -4
Lim √x+4 = √( -4+4) = 0
x→ -4
2. Lim (√x-√a)/(x-a) = (√a-√a)/(a-a) =0/0 ind
x→ a
Lim (√x-√a)/(x-a) * (√x+√a)/(√x+√a) =Lim (√x )²- (√a )² /( x-a)*(√x +√a )
x→ a x→ a
= Lim (x-a )/(x-a)* (√x+√a) = Lim 1/(√x+√a) = 1/(√a +√a ) = 1/2√a
x→ a x→ a
= 1/2√a*√a/√a = √a/2a
3. Lim x+1 /√6x²+3 +3x = (-1+1)/(√6*(-1)²+3 +3*(-1))=0/0
x→-1
Lim( x+1 )/(√6x²+3 +3x)*(√6x²+3 -3x)/(√6x²+3 -3x)=
x→-1
Lim ( x+1 )*(√6x²+3 -3x)/(√6x²+3)² -(3x)² =
x→-1
Lim ( x+1 )*(√6x²+3 -3x)/( 6x²+3 -9x²) =Lim( x+1 )*(√6x²+3 -3x)/( -3x²+3)
x→-1 x→-1
Lim( x+1 )*(√6x²+3 -3x)/-3*(x+1 )(x-1 ) = Lim (√6x²+3 -3x)/-3*(x-1 ) =
x→-1 x→-1
= (√6*(-1)²+3 -3*(-1))/-3*(-1-1 ) = 6/6 = 1
4. Lim (√x+2 - 2) /(x-2) = 0/0
x→2
Lim (√x+2 - 2) /(x-2) =Lim (√x+2 - 2)*(√x+2 + 2) /(x-2)* (√x+2 + 2)=
x→2 x→2
Lim (√x+2)²- (2)²/(x-2)* (√x+2 + 2)= Lim (x-2)/(x-2)* (√x+2 + 2)
x→2 x→2
= Lim 1/(√x+2 + 2) = 1/(√2+2 + 2 ) = 1/(√4 + 2 ) = 1/(2 +2 ) = 1/4
x→2
5. Lim( √x²+25 - 5) /(√x² +16 -4 ) =
x→0
=Lim( √x²+25 - 5)* (√x² +16 +4 )*( √x²+25 +5) /(√x² +16 -4 )*(√x² +16 +4 )*( √x²+25 +5)
= Lim ( √x²+25)² - (5)²)*(√x² +16 +4 )/(√x² +16)² -(4 )²*( √x²+25 +5) =
x→0
= Lim x²*(√x² +16 +4 )/x²*(√x²+25 +5) =Lim(√x² +16 +4 )/(√x²+25 +5)
x→0 x→0
= ( √0²+16 + 4 )/(√0²+25 + 5 ) = 8/10 = 4/5