Trabajo Colaborativo 2 Calculo

FASE 1

Resuelva los siguientes límites:

〖lim〗_(a→π) 4cos3a – 2sen2a

=4 cos⁡(3×180°)-2sen(2×180°

=4cos540°-2sen360°

=-4-0

=-4

〖lim〗_(a→1) √(x^2+3x)-√(x^2+x)

=√((1)^2+3(1) )-√((1)^2+1)

=√4-√2

=2-√2

Demuestre que:

〖lim〗_(a→0) ((x+h)^3-x^3)/h=3x^2

=〖lim〗_(a→0) (x^3+3xh^2+3xh^2+h^3-x^3)/h = (h(3x^2+3xh+h^3))/h

=3x^2+3x(0)+0^2=5x^2+0+0

=3x^2

FASE 2

Halle los siguientes límites infinitos

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

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

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

=〖lim〗_(a→∞) (-a^2-9a-19)/(a^2+3a+2)

=(a^2/a^2 -9a/a^2 -19/a^2 )/(a^2/a^2 +3^2/a^2 +2/a^2 ) = (-1-9/a-19/a^2 )/(1+3/a+2/a^2 ) = (-1-0-0)/(1+0+0)

=(-1)/1

=-1

5. 〖lim〗┬(U→0)⁡〖(sin⁡2 (U/2))/U^2 〗

lim┬(U→0)⁡〖(sin⁡2 (U/2))/U^2 〗 〖 → 〖 lim〗┬(U→0)〗⁡〖(sin⁡2 (0/2))/0^2 〗

lim┬(U→0)⁡〖[sin⁡2 (U/2) ]^2/(U^2*1/2)〗

lim┬(U→0)⁡√(1/2)*lim┬(U→0) ( sin⁡2 (U/2))/U^2 U/2=X

1/4*lim┬(U→0) sin⁡X/X

1/4*(1)=1/4

6. 〖lim⁡〗┬(X→0) tan⁡4X/(sin⁡2 X)

〖lim⁡〗┬(X→0) sin⁡4X/(cos⁡4 X)*〖lim⁡〗┬(X→0) 1/sin⁡2X

〖lim⁡〗┬(X→0) 2sin⁡4X/(cos⁡4 X)*〖lim⁡〗┬(X→0) 1/cos⁡4X *〖lim⁡〗┬(X→0) 1/sin⁡2X

4X=a

〖2lim⁡〗┬(X→0)

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