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प्रश्न
Evaluate the following limits:
`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`
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उत्तर
`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1) = lim_(x -> 1) (root(3)(7 + x^2) - 2 + 2 sqrt(3 + x^2))/(x - 1)`
= `lim_(x -> 1) ((7 + x^3)^(1/3) - 2)/(x - 1) - lim_(x -> 1) ((3 + x^2)^(1/2) - 2)/(x - 1)`
= `lim_(x -> 1) ((7 + x^3)^(1/3) - (8)^(1/3))/(x^3 - 1) xx (x^3 - 1)/(x - 1) - lim_(x -> 1) ((3 + x^2)^(1/2) - (4)^(1/2))/(x^2 - 1) xx (x^2 - 1)/(x - 1)`
= `lim_(x -> 1) ((7 + x^3)^(1/3) - (8)^(1/2))/((7 + x^3) - 8) xx ((x - 1)(x^2 + x + 1))/(x - 1) - lim_(x -> 0) ((3 + x^2)^(1/2) - (4)^(1/2))/((3 + x^2) - 4) xx ((x + 1)(x - 1))/(x - 1)`
= `lim_(x -> 1) ((7 + x^3)^(1/3) - (8)^(1/2))/((7 + x^3) - 8) xx (x^2 + x + 1) - lim_(x _> 1) ((3 + x^2)^(4)^(1/2))/((3 + x^2) - 4) xx (x + 1)`
`lim_(x -> "a") (x^"n" - "a"^"n") = "na"^("n" - 1)`
= `1/3(8)^(1/3 - 1) (1^2 + 1 + 1) - 1/2(4)^(1/2 - 1) (1 + 1)`
= `1/3(8)^(-2/3) (3) - 1/2 xx (4)^(-1/2) xx (2)`
= `(2^3)^(-2/3) - (2^2)^(- 1/2)`
= `2^(-2) - 2^(-1)`
= `1/2^2 - 1/2`
= `1/4 - 1/2`
= `(1 - 2)/4`
= `- 1/4`
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