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Question
Evaluate the following: `lim_(x -> 0) [(2^x - 1)^2/((3^x - 1) xx log (1 + x))]`
Solution
`lim_(x -> 0) [(2^x - 1)^2/((3^x - 1) xx log (1 + x))]`
= `lim_(x -> 0) ((2^x - 1)^2/x^2)/((3^x - 1*log 1 + x)/x^2) ...[("Divide Numerator and"),("Denominator by" x^2),("As" x -> 0"," x ≠ 0),(therefore x^2 ≠ 0)]`
= `(lim_(x -> 0) ((2^x - 1)/x)^2)/(lim_(x -> 0) [((3^x - 1)/x) xx (log 1 + x)/x]`
= `(lim_(x -> 0) ((2^x - 1)/x)^2)/(lim_(x -> 0) ((3^x - 1)/x) xx lim_(x -> 0) (log 1 + x)/x)`
= `(log 2)^2/(log 3 xx 1) ...[(lim_(x -> 0) ("a"^x - 1)/x = log"a"","),(lim_(x -> 0) (log(1 + x))/x = 1)]`
= `(log 2)^2/log3`
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