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Evaluate the following limits: limx→2[x3-8x+2-3x-2] - Mathematics and Statistics

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प्रश्न

Evaluate the following limits: `lim_(x -> 2) [(x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2))]`

योग

उत्तर

`lim_(x -> 2) [(x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2))]`

= `lim_(x -> 2) [(x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2)) xx (sqrt(x + 2) + sqrt(3x - 2))/(sqrt(x + 2) + sqrt(3x - 2))]`

= `lim_(x -> 2) ((x^3 - 8)(sqrt(x + 2) + sqrt(3x - 2)))/((x + 2) - (3x - 2)`

= `lim_(x -> 2) ((x^3 - 8)(sqrt(x + 2) + sqrt(3x - 2)))/(-2x + 4)`

= `lim_(x -> 2) ((x - 2)(x^2 + 2x + 4)(sqrt(x + 2) + sqrt(3x - 2)))/(-2(x - 2))`

= `lim_(x -> 2) ((x^2 + 2x + 4)(sqrt(x + 2) + sqrt(3x - 2)))/(-2)      ...[("As"  x -> 2","  x ≠ 2),(therefore x - 2 ≠)]`

= `-1/2 lim_(x -> 2) (x^2 + 2x + 4) xx lim_(x -> 2) (sqrt(x + 2) + sqrt(3x - 2))`

= `-1/2 xx [(2)^2 + 2(2) + 4] xx (sqrt(2 + 2) + sqrt(3(2) - 2))`

= `-1/2 xx 12 xx (2 + 2)`

= – 6 x 4
= – 24

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अध्याय 7: Limits - EXERCISE 7.3 [पृष्ठ १०३]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] 11 Standard Maharashtra State Board
अध्याय 7 Limits
EXERCISE 7.3 | Q III. 4. | पृष्ठ १०३

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