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Select the correct answer from the given alternatives. limx→0[log(5+x)-log(5-x)sinx] =

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प्रश्न

Select the correct answer from the given alternatives.

`lim_(x -> 0) [(log(5 + x) - log(5 - x))/sinx]` =

पर्याय

  • `3/2`

  • `-5/2`

  • `-1/2`

  • `2/5`

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उत्तर

`2/5`

Explanation;

`lim_(x -> 0) [(log(5 + x) - log(5 - x))/sinx]`

= `lim_(x -> 0)(log[5(1 + x/5)] - log[5(1 - x/5)])/sinx`

= `lim_(x -> 0) (log5 + log(1 + x/5) - [log5 + log(1 - x/5)])/sinx`

= `lim_(x -> 0)[(log(1 + x/5) -log(1 - x/5))/x xx x/sinx]`

= `lim_(x -> 0) [log(1 + x/5)/(5(x/5)) - (log(1 - x/5))/((-5)((-x)/5))]  xx  lim_(x -> 0) x/sinx`

= `[1/5 (1) + 1/5(1)] xx 1`

= `2/5`

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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. (10) | पृष्ठ १५८

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


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