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Evaluate the following: limx→0[log(2+x)-log(2-x)x]

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प्रश्न

Evaluate the following: `lim_(x -> 0)[(log(2 + x) - log( 2 - x))/x]`

योग
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उत्तर

`lim_(x -> 0)[(log(2 + x) - log( 2 - x))/x]`

= `lim_(x -> 0)(log[2(1 + x/2)] - log[2(1 - x/2)])/x`

= `lim_(x -> 0) (log2 + log(1 + x/2) - [log2 + log(1 - x/2)])/x`

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

= `lim_(x -> 0) [(log(1 + x/2))/x - (log(1 - x/2))/x]`

= `lim_(x -> 0) [(log(1 + x/2))/(2(x/2)) - (log(1 - x/2))/((-2)(-x/2))]`

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

= `1/2(1) + 1/2(1)    ....[(because x -> 0"," x/2 -> 0 and),(lim_(x -> 0) (log(1 + x))/x = 1)]`

= 1

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अध्याय 7: Limits - EXERCISE 7.4 [पृष्ठ १०५]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
EXERCISE 7.4 | Q I] 3) | पृष्ठ १०५

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