Question

Límites y continuidad

Calcular los siguientes límites:

Answer 1

SOLUCIÓN :

1 ) lim v→ 3 √v+1  - 2 / v -3 = √3+1 - 2 /3-3 =0/0

     lim v→3 ( √v+1 - 2 )(√v+1  + 2 )/(v-3 )(√v+1 + 2 )=

         lim v→3 (v-3)/(v-3)*(√v+1 +2 ) =

          lim v→3  1/(√v+1 +2) = 1/(√3+1 + 2 ) = 1/4 .

   2 ) lim x→1  (x³- 1 )/(x²- 1 )= 1³-1/1²-1 =0/0

        lim x→1  ( x-1 )(x²+ x +1 )/( x-1 )(x+1 ) =

        lim x→1 ( x²+ x+1 )/( x+1 )=

         =  1²+1+1 /1+1= 3/2

   3 ) lim x→∞ ( 3x -√4x+2 ) = ∞-∞

         lim x→∞  ( 3x - √4x+2  )*( 3x + √4x+2  )/( 3x + √4x+2  )=

         lim x→∞ (9x²- 4x - 2)/( 3x + √4x+2  )= ∞/∞

          lim x→∞  ( 9x²/x² - 4x/x²-2/x²)/( 3x/x²+ √( 4x/x⁴+ 2/x⁴)

         lim x→∞  ( 9 - 4/x - 2/x²)/( 3/x + √( 4/x³+ 2/x⁴ ) = 9/0 = ∞

   4 ) lim x→0   1/x * senx/3 = 1/0 *sen 0/3 = ind.

         lim x→0  sen(x/3)/x = lim x→0 sen(x/3)/3x/3 =

           = 1/3 * lim x→0 sen(3/x)/(x/3) = 1/3 * 1 = 1/3 .

          como lim x→0 senx/x = 1 .