ACT 1 CALCULO DIFERECIAL

Limites 1 parte

demostrar que el límite de la función 〖(4〗⁡x+1) cuando x tiende a 2 es 9

resolver los siguientes limites:

lim⁡x→(-2) [(x^2-4)/(x-2)]=[((-2)^2-4)/(-2-2)]=[(-4-4)/(-4)=] (-8)/(-4)=2

b. lim⁡x→0 [〖2e〗^2x+(x-6)/(x+1)]= [〖2e〗^(2(0))+(0-6)/(0+1)]= [2e+(-6)/1]= [2e+(-6)]

si lim⁡x→a [f(x)]=3 entonces:

a. Hallar lim⁡x→a [f(x) ]^4=?

lim⁡x→a [3]^4=81

b. Hallar lim⁡x→a [3 f(x)-2]=lim⁡x→a[3 f(a)-2]

4. sea f(x)=Ln|x| y g(x)=√(x^2 )-3 hallar lim⁡x→2 [fog](x)

lim⁡x→2 [fog](x)=?

lim⁡x→2[√(x^2 )-3=lim⁡x→2 [√4-3]=1

Limites 2 parte

Resolver los siguientes limites:

lim⁡x→∞ [(〖4x〗^3-2x+1)/(〖2x〗^3+5x-9)]

lim⁡x→∞ [(〖6x〗^3+〖10x〗^2-3)/(〖3x〗^4+〖2x〗^2-6)]

lim⁡x→∞ [(〖4x〗^3-〖5x〗^2+8x-3)/(〖2x〗^2+8x-6)]

lim⁡x→∞[(3√(x^4+x+2) + 5√(x^3+3x^2 ) +x+1)/(4√(x^6+3x+2 +) 5√(x^2+4x+7))]

lim⁡x→∞ √x*[√(x+3) -√(x+2)

Resolver:〖 lim〗⁡h→0[((x+h)^3-x^3)/h]=

(〖x+h)〗^3=x^3+3xh+h^3

(〖x+h)〗^3-x^3=3xh+h^3

((〖x+h)〗^3-x^3 ))/h=(3xh+h^3)/h=3x+h

Ahora está claro que el límite de 3x+h cuando h tiende a 0 es 3x

Resolver: lim⁡x→1[(x^4+x^3+x^2+x-4)/(x-1)]=[(1+1+1+1-4)/(1-1)]=0/0 indeterminacion

Resolver: lim⁡x→0 [(√(a+x) –√(a-x))/x ]=lim⁡x→0 [((√a+0)-(√a-0))/0 ]=0/0 indeterminacion

lim⁡x→0 [(√a+x-√a-x)/x ]=lim⁡x→0 [((√a+x-√a-x))/(x(√a+x-√a-x)) ]=lim⁡x→0 [((a-x))/(x(√a+x-√a-x)) ]=[((x))/(x(√a+x-√a-x)) ]= [((1))/((√a+x-√a-x)) ]=1/(2√x)

Resolver: lim⁡x→0 [(sen(8x)+sen(4x))/(sen 6x)]=(sen(0)+sen(0))/(sen 6(0))=2sen/(sen )

6. lim⁡x→∞ (1+3/x〖) 〗^(5/x)=(1+3/∞)=(1+((1+∞))/∞)=∞/∞

7. a. lim⁡x→0 [(1-cos⁡〖(X)〗)/x^2 ]=((1-cos⁡〖(X)) (1+cos⁡(x))〗)/(〖[x〗^2*((1+cos⁡(x) )])=((1-cos⁡〖(X))〗)/(〖[x〗^2*((1+cos⁡(x) )])

=(sen^2 (x))/(〖[x〗^2*((1+cos⁡(x) )])=(sen (x))/((x) )*(sen (x))/((x) )=1*1*(1/2)=1/2

b. lim⁡x→0 [(1-cos⁡〖(2X)〗)/(sen^2 (2x))]

c. lim⁡x→π[(tan^2 (x))/(1+cos⁡(x))]

8.

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