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Question
Evaluate the following limits:
`lim_(x -> 0) (2^x - 3^x)/x`
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Solution
We know `lim_(x -> 0) ("a"^x - 1)/x = log "a", "a" > 0`
`lim_(x -> 0) (2^x - 3^x)/x = lim_(x -> 0) (2^x - 1 + 1 - 3^x)/x`
= `lim_(x -> 0) ((2^x - 1) - (3^x - 1))/x`
= `lim_(x -> 0) ((2^x - 1)/x - (3^x - 1)/x)`
= `lim_(x -> 0) ((2^x - 1)/x) - lim_(x -> 0) ((3^x - 1)/x)`
= log 2 – log 3
`lim_(x -> 0) (2^x - 3^x)/x = log 2/3`
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