Anónimo
Anónimo preguntado en Ciencias y matemáticasMatemáticas · hace 4 meses

¿Quien puede resolver este limite ?

Lim (x^2-4)/(x^2+4) cuando x -->2

si el numerador es 0 entonces sigue siendo forma indeterminada? como seria el resultado ? Alguien me puede explicar porfa

gracias de antemano

7 respuestas

Calificación
  • hace 4 meses

    Es cero en el numerador 

    El resultado es cero no es indeterminado

  • hace 4 meses

    Hay diversas indeterminaciones, cuando es de la forma x/0 tiende a mas infinito o menos infinito dependiendo el caso, para el que expones el limite si existe y es 0, dado que no existe una indeterminación.

    PD: El resultado se obtiene al sustituir el valor al cual tiende la variable "x"

  • Anónimo
    hace 3 meses

    Lim (x^2-4)/(x^2+4) = 0/8 = 0

    x->2, el numerador es cero ;)

  • Anónimo
    hace 3 meses

     si , El numerador es cero, si se puede dividir cero entre algo, lo que no se puede es dividir algo entre cero.

    Lim (x^2-4)/(x^2+4) = 0/8 = 0

    x->2

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  • Anónimo
    hace 3 meses

    El numerador es cero, si se puede dividir cero entre algo, lo que no se puede es dividir algo entre cero.

    Lim (x^2-4)/(x^2+4) = 0/8 = 0

    x->2

  • Anónimo
    hace 4 semanas

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  • hace 4 meses

    parte

    I ante una fracción racional hallas el "conjunto restricción"

     igualando a cero el denominador. Y resuelves esa ecuación

    x^2 + 4 = 0

    x^2 = - 4

    x = 2i

    no existe restricción  dentro del os números reales

    observación = no acemos restrcción al numerador, sólo al denominador

    si el denominador es distinto de cero, podemos reemplazar libremente. SI además el numerador es cero entonces el resultado será cero

    Parte II

    supongamos que buscas determinar el límite de la siguiente función

    Lim (x^2+4) / (x^2-4) cuando x -->2

    primero hacemos la restricción

    x^2 - 4 = 0x^2 = 4x1 = +2

    x2 = - 2en estos dos valores se puede indefinir, por lo tanto no podemos hacer una reemplazo directo, pero igualmente puede ser alguno de esos valores, el valor del límite

    como   x -->2  lo ideal es probar número menores a 2 muy cercanos a 2. Asimismo números mayores a 2 , muy cercanos a 2. Analizams a qué número tienden cuando los reemplazamos

    con los números mayores, obtenemos el límite por la derecha:

    Lim (1.7^2+4) / (1,7^2-4) cuando x -->2    = -6,207207207

    Lim (1,8^2+4) / (1,8^2-4) cuando x -->2 = -9,526315789

    Lim (1,9^2+4) / (1,9^2-4) cuando x -->2 = -19,51282051

    para 1,99 el valor de la funcipon es -199,5012531

    para 1,999 el valor de la función es -1999,500125

    para 1,9999  el valor de la función es  19999,5

    para 1,99999 = -199999,5

    para x = 1,999999    el valor de la función es = - 1999999.5

    con los números menores  obtenemos el límite por la izquierda

    para,2,5 el valor de la función es 4,5555555

    para 2,4 el valor será 5,45454545

    para  2,3 el valor es 7,201550388

    para 2,2 el valor es 10,52380952

    para 2,1 el valor es de 20,51219512

    para 2,11   el valor 18,69520017

    para 2,111 el valor es de 18,53151838

    para 2,1111 el valor 18,51531237

    para 2,11111 el valor 18,51369338

    para 2,111111111111 el vlor de la función es 18,51351353

    al parecer el límite por la derecha tiende a 18,51

    pero el límite por la izquierda no tiene a un número definido, sino a 1999....999

    por ende no existe el límite por la izquierda

    como no se cumple que seam iguales lel ímite por la derecha y el límite por la izquierda, concluimos que no existe el límite cuando x iende a 2 de la función dada

    observación: de probar este método de acercamiento con el método anterior te hubiere dado el valor de lso 3 límites

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