Trabajo cálculo Diferencial

FASE 1

lim┬(n→ -1)⁡〖(√(5+n)-2)/(n+1)〗

lim┬(n→ -1)⁡〖((√(5+n)-2)(√(5+n)+2))/((n+1)(√(5+n)+2))〗

lim┬(n→ -1)⁡〖(5+n-4)/((n+1)(√(5+n)+2))〗

lim┬(n→ -1)⁡〖((n-1))/((n+1)(√(5+n)+2))〗

lim┬(n→ -1)⁡〖1/((√(5+n)+2))〗

lim┬(n→ -1)⁡〖1/((√(5-1)+2))〗

lim┬(n→ -1)⁡〖1/((√4+2))〗

lim┬(n→ -1)⁡〖1/((2+2))〗

〖lim┬(n→ -1)= 〗⁡〖1/4〗

lim┬(a→ π)⁡〖2cos2a-4sen3a〗

lim┬(a→ π)⁡〖= 2cos(2π)-4sen(3π)〗

lim┬(a→ π)=2-0

lim┬(a→ π)=2

lim┬(x→ 1)⁡〖√(x^2+3x)- √(x^2+x)〗

lim┬(x→ 1)⁡〖√(1^2+3*1)- √(1^2+1)〗

lim┬(x→ 1)⁡〖√4- √2〗

〖lim┬(x→ 1)= 〗⁡〖2- √2〗

lim┬(x→ 1)=-2√2

Demuestre que:

lim┬(h→ b)⁡〖(〖(b+h) 〗^2- b^2)/h=3b〗

lim┬(h→ b)⁡〖(〖(b+h) 〗^2- b^2)/h=3b〗

lim┬(h→ b)⁡〖(b^2+2bh+ h^2- b^2)/h=3b〗

lim┬(h→ b)⁡〖(h^2+2bh)/h=3b〗

lim┬(h→ b)⁡〖(h(h+2b))/h=3b〗

lim┬(h→ b)⁡〖h+2b=3b〗

lim┬(h→ b)⁡〖b+2b=3b〗

lim┬(h→ b)⁡〖3b=3b〗

5. lim┬(h→ 0)⁡〖(〖(x+h) 〗^3- x^3)/h=〖3x〗^2 〗

lim┬(h→ 0)⁡〖(x^3+3hx^2+〖3h〗^2 x+h^3- x^3)/h=〖3x〗^2 〗

lim┬(h→ 0)⁡〖(3hx^2+〖3h〗^2 x+h^3)/h=〖3x〗^2 〗

lim┬(h→ 0)⁡〖(h(3x^2+3hx+h^2))/h=〖3x〗^2 〗

lim┬(h→ 0)⁡〖3x^2+3hx+h^2=〖3x〗^2 〗

lim┬(h→ 0)⁡〖3x^2+3(0)x+〖(0)〗^2=〖3x〗^2 〗

lim┬(h→ 0)⁡〖3x^2+0+0=〖3x〗^2 〗

lim┬(h→ 0)⁡〖3x^2=〖3x〗^2 〗

FASE 2

C. Demuestre los siguientes límites infinitos:

6)

〖lim〗_(a→∞) {(a^2+1)/(a+2)-(a^2+10)/(a+1)}=-1

〖lim〗_(a→∞) (〖(a〗^2+1)(a+1)-(a+2)(a^2+10))/(a+2)(a+1) = -1

〖lim〗_(a→∞ ) ((a^3+a^2+a+1-a^3+10a+〖2a〗^2+20))/(a^2+a+2a+2)=-1

〖lim〗_(a→∞ ) (〖-a〗^2+11a+21)/(a^2+3a+2)= -1

〖lim〗_(a→∞ ) (〖-a〗^2/a^2 +11a/a^2 +21/a^2 )/(a^2/a^2 +(3a/a^2 )+(2/a^2 ) )=-1

〖lim〗_(a→∞) -1/1=-1

〖lim〗_(a→∞ ) -1=-1

7) lim┬(a → ∞)⁡〖√(x^2+x)-x〗= 1/2

lim┬(a → ∞)⁡〖(√(x^2+x)-x)* √(x^2+x)/√(x^2+x)〗

lim┬(a → ∞)⁡〖(〖√(x^2+x)〗^2-x^2)/(√(x^2+x)+x)〗

lim┬(a → ∞)⁡〖(〖√(x^2+x)〗^2-x^2)/(√(x^2+x)+x)〗

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