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Evaluate the following limit : limx→0[(25)x-2(5)x+1x⋅sinx]

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प्रश्न

Evaluate the following limit : 

`lim_(x -> 0) [((25)^x - 2(5)^x + 1)/(x*sinx)]`

योग
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उत्तर

`lim_(x -> 0) ((25)^x - 2(5)^x + 1)/(x*sinx)`

= `lim_(x -> 0) ((5^x)^2 - 2(5^x) + 1)/(x*sinx)  ...[(25)^x = (5^2)^x = (5^x)^2]`

= `lim_(x -> 0) ((5^x - 1)^2)/(x*sinx)`

= `lim_(x -> 0) ((5^x - 1)^2/x^2)/((x*sinx)/x^2)  ...[("Divide numerator and"),("denominator by"  x^2.),(because x -> 0","  x ≠ 0),(therefore x^2 ≠ 0)]`

= `(lim_(x -> 0)((5^x - 1)/x)^2)/(lim_(x -> 0) sinx/x)`

= `(log5)^2/1  ...[∵ lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

= (log 5)2

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अध्याय 7: Limits - Exercise 7.6 [पृष्ठ १५४]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Exercise 7.6 | Q III. (4) | पृष्ठ १५४

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