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Evaluate the following limit : limx→0[6x+5x+4x-3x+1sinx]

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Question

Evaluate the following limit : 

`lim_(x -> 0) [(6^x + 5^x + 4^x - 3^(x + 1))/sinx]`

Sum
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Solution

`lim_(x -> 0) [(6^x + 5^x + 4^x - 3^(x + 1))/sinx]`

= `lim_(x -> 0) [((6^x - 1) + (5^x - 1) + (4^x - 1) - 3^(x + 1) + 3)/sinx]`

= `lim_(x -> 0) ((6^x - 1) + (5^x - 1) + (4^x - 1) - 3(3^x - 1))/sinx`

= `lim_(x -> 0) (((6^x - 1)/x) + ((5^x - 1)/x) + ((4^x - 1)/x) - 3((3^x - 1)/x))/((sinx/x)`   ...[∵ x → 0 ∴ x ≠ 0]

= `(lim_(x -> 0) (6^x - 1)/x + lim_(x -> 0) (5^x - 1)/x + lim_(x -> 0) (4^x - 1)/x - 3 lim_(x -> 0)(3^x - 1)/x)/((lim_(x -> 0) sinx/x))`

= `(log 6 + log 5 + log 4 - 3 log 3)/1   ...[because lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

= log(6 × 5 × 4) – log 33 

= `log((6 xx 5 xx 4)/27)`

= `log(40/9)`.

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Chapter 7: Limits - Exercise 7.6 [Page 154]

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