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Question
Evaluate the following: `lim_(x -> 0) [("a"^(3x) - "b"^(2x))/(log 1 + 4x)]`
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Solution
`lim_(x -> 0) [("a"^(3x) - "b"^(2x))/(log 1 + 4x)]`
= `lim_(x -> 0) ("a"^(3x) - 1 - "b"^(2x) - 1)/(log 1 + 4x)`
= `lim_(x -> 0) [("a"^(3x) - 1 - "b"^(2x) - 1)/x)/((log 1 + 4x)/x)`
= `(lim_(x -> 0) [("a"^(3x) - 1)/x - ("b"^(2x) - 1)/x])/(lim_(x -> 0) (log 1 + 4x)/(x)`
= `(lim_(x -> 0)[("a"^(3x) - 1)/(3x)] xx 3 - lim_(x -> 0)[("b"^(2x) - 1)/(2x)] xx 2)/(lim_(x -> 0) (log 1 + 4x)/(4x) xx 4)`
= `(3log "a" - 2 log"b")/(1 xx 4) ...[(because x -> 0"," 2x -> 0"," 3x -> 0),(4x -> 0 and lim_(x -> 0) ("a"^x - 1)/x = log "a"),(and lim_(x -> 0) (log (1 + x))/x = 1)]`
= `1/4(log"a"^3 - log"b"^2)`
= `1/4 log("a"^3/"b"^2)`
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