Ejercicios de Matemáticas

FASE 1

A . resuelve los siguientes límites:

〖lim〗_(x →-2) (x^2-x-2)/(x^2-5x+6)

lim (x^2-x-2)/(x^2-5x+6) - lim (( x-2)(x+1))/((x-2)(x-3))

x →2

= lim (x+1)/(x-3)

x →2

= lim x+1

x →2

lim x-3

x →2

= 3/(-1)

= -3

2. 〖lim〗_(x →0)

lim √(9+x -3)/x = lim √(9+x -3)/x √(9+x+3)/√(9+x+3)

x →0 x →0

= lim ((√(9+x)〖2-〗_9 ))/x(√(9+x)+3)

= lim

x →0 (9+x-9)/(x(√(9+x)+3))

= lim

x →0 1/√(9+x +3)

= 1/√(9+0+3 ) = 1/6

3. 〖lim〗_(x →-2) (3-√(x^2+5))/(3x+6)

(3-√(x^2+5))/(3x+6) = lim

x →-2 x →-2 (3-√(x^2+5))/(3x+6) x (3+√(x^2+5))/(3+√(x^2+5))

= lim (9-(x^2+5))/(9x+3x√(x^2+5 )) +18+6√(x^2+5))

x →-2

= lim

x →-2 (4-x^2)/(9 (x+2)+3√(x^2+5(x+2)))

= lim

x →-2 ((2-x)(2+x))/((x+2)(9+3√(x^2+5))

= lim

x →-2 (2-x)/(9+3√(x^2+5))

= lim

(x →-2 2-x)/█(lim 9+3√(x^2+5)@x →-2) = (2-(-2))/(9+3√(4+5))

=4/18 = 2/9

4. 〖lim〗┬h⁡〖→2b〗 ((b+h)2_(-b^2 ))/h→

((b+h)2_(-b^2 ))/h = ((b+2b)2_(-b^2 ))/2b

= ((3〖b)〗^2 )- b^2)/2b = (9b^2-b^2)/2b = 4b

FASE 3

X3/(1-2X^2 )

8. 〖lim〗_(x →00) {X^3/〖4X〗^3 }

Simplificando se tiene:

〖lim〗_(x →00) {1/4} x^3⁄(1-〖2x〗^3 ) = 〖lim〗_(x →00) {1/4} (x^2⁄x^3 )/((1-2x^3 )/x^3 )

...