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Question
Evaluate the following :
`lim_(x -> 0) [("a"^(3x) - "a"^(2x) - "a"^x + 1)/(x*tanx)]`
Solution
`lim_(x -> 0) [("a"^(3x) - "a"^(2x) - "a"^x + 1)/(x*tanx)]`
= `lim_(x -> 0) ("a"^(2x) * "a"^x - "a"^(2x) - "a"^x + 1)/(xtanx)`
= `lim_(x -> 0) ("a"^(2x)("a"^x - 1) - ("a"^x - 1))/(xtanx)`
= `lim_(x -> 0) (("a"^x - 1)("a"^(2x) - 1))/(xtanx)`
= `lim_(x -> 0)((("a"^x - 1)/x)(("a"^(2x) - 1)/x))/((tanx/x)` ...[∵ x → 0, ∴ x ≠ 0]
= `((lim_(x -> 0) ("a"^x - 1)/x) xx (lim_(x -> 0) (("a"^2)^x - 1)/x))/((lim_(x -> 0) tanx/x))`
= `((log "a")(log "a"^2))/1 ...[because lim_(x -> 0) ("a"^x - 1)/x = log "a"]`
= (log a)(2 log a)
= 2(log a)2
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