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Select the correct answer from the given alternatives. limx→0[(3sinx-1)3(3x-1).tanx.log(1+x)] =

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प्रश्न

Select the correct answer from the given alternatives.

`lim_(x→0)[(3^(sinx) - 1)^3/((3^x - 1).tan x.log(1 + x))]` =

विकल्प

  • 3log 3

  • 2log 3

  • (log 3)2 

  • (log 3)3 

MCQ
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उत्तर

`lim_(x→0)[(3^(sinx) − 1)^3/((3^x − 1).tan x.log(1 + x))]` = (log 3)2 

Explanation:

`lim_(x→0)[(3^(sinx) - 1)^3/((3^x - 1).tan x.log(1 + x))]`

`= (lim_(x →0) (3^(sin x) - 1)^3/(sin^3 x). (sin^3 x)/(x^3))/(lim_(x →0) ((3^x − 1)/x).(tan x/x).log(1 + x)/x)`

`= (lim_(x →0) ((3^(sinx) - 1)/sin x)^3. lim_(x →0) (sin x/x)^3)/(lim_(x →0) ((3^x − 1)/x). lim_(x →0) (tan x/x). lim_(x →0) (log(1 + x)/x))`

`= ((log 3)^3 × (1)^3)/(log 3 × 1 × 1)`

`= (log 3)^3/(log 3)`

= (log 3)2

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अध्याय 7: Limits - Miscellaneous Exercise 7.1 [पृष्ठ १५८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Miscellaneous Exercise 7.1 | Q I. (13) | पृष्ठ १५८

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