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Select the correct answer from the given alternatives. limx→0[x⋅log(1+3x)(e3x-1)2] =

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प्रश्न

Select the correct answer from the given alternatives.

`lim_(x -> 0) [(x*log(1 + 3x))/("e"^(3x) - 1)^2]` =

पर्याय

  • `1/"e"^9`

  • `1/"e"^3`

  • `1/9`

  • `1/3`

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उत्तर

`1/3`

Explanation;

`lim_(x -> 0) (x*log(1 + 3x))/("e"^(3x) - 1)^2` 

= `(lim_(x -> 0) (log(1 + 3x))/x)/(lim_(x -> 0)((e^(3x) - 1)/x)^2`

= `(lim_(x -> 0) [(log(1 + 3x))/(3x) xx 3])/(lim_(x -> 0)[(("e"^(3x) - 1)/(3x))^2 xx (3)^2]`

= `1/3`

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

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
पाठ 7 Limits
Miscellaneous Exercise 7.1 | Q I. (12) | पृष्ठ १५८

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\[\lim_{x\to0}\frac{\mathrm{e}^{\tan x}-\mathrm{e}^{x}}{\tan x-x}=\]


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