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Evaluate the following limit : limy→1[2y-27+y3-2]

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प्रश्न

Evaluate the following limit :

`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 2)]`

योग
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उत्तर

`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y )- 2)]`

= `lim_(y -> 1) [(2y - 2)/((7 + y)^(1/3) - 8^(1/3))] xx [((7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3))/((7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3))]`

= `lim_(y -> 1) ((2y - 2)[(7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3)])/((7 + y) - 8)   ...[because "a" - "b" = ("a"^(1/3) - "b"^(1/3)) ("a"^(2/3) + "a"^(1/3)"b"^(1/3) + "b"^(2/3))]`

= `lim_(y -> 1) (2(y - 1)[(7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3)])/(y - 1)`

= `lim_(y -> 1) 2[(7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3)]   ...[(because y -> 1","  y ≠ 1","),(therefore y - 1 ≠ 0)]`

= `2[(7 + 1)^(2/3) + (7 + 1)^(1/3) 8^(1/3) + 8^(2/3)]`

= = 2[4 + 4 + 4] ...`[∵ 8^(1/3) = 2 and 8^(2/3) = 4]`

= 24

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Exercise 7.1 [पृष्ठ १३९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Exercise 7.1 | Q III. (6) | पृष्ठ १३९

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