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Evaluate the following Limits: limx→0[x(6x-3x)(2x-1)⋅log(1+x)]

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प्रश्न

Evaluate the following Limits: `lim_(x -> 0)[(x(6^x - 3^x))/((2^x - 1)*log(1 + x))]`

योग
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उत्तर

`lim_(x -> 0)(x(6^x - 3^x))/((2^x - 1)*log(1 + x))`

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

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

= `lim_(x -> 0) (x*3^x)/(log (1 + x))    ...[("As"  x -> 0","  2^x -> 2^0),("i.e." 2^x -> 1  therefore 2^x ≠ 1),(therefore 2^x - 1 ≠ 0)]`

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

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

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

= 1

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अध्याय 7: Limits - MISCELLANEOUS EXERCISE - 7 [पृष्ठ १०६]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
MISCELLANEOUS EXERCISE - 7 | Q II. 13) | पृष्ठ १०६

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