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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Evaluate the following limits: limx→22-x+223-4-x3

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प्रश्न

Evaluate the following limits:

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

बेरीज
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उत्तर

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

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

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

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

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

`lim_(x -> "a") (x^"n" - "a"^"n") = "na"^("n" - 1)`

= `1/2(4)^(1/2 - 1) xx - 1/(1/3 (2)^(1/3 - 1)`

= `1/2(4)^(-1/2) xx - 3/((2)^(-2/3)`

= `1/(2(2^2)^(1/2)) xx - 3 xx 2^(2/3)`

= `- 1/(2 xx 2) xx 3 xx 2^(2/3)`

= ` - 3/4 xx (2^2)^(1/3)`

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

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पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.2 [पृष्ठ १०३]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.2 | Q 11 | पृष्ठ १०३

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