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Question
Evaluate the following: `lim_(x -> 0)[(log(3 - x) - log(3 + x))/x]`
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Solution
`lim_(x -> 0)(log(3 - x) - log(3 + x))/x`
= `lim_(x -> 0) 1/x log ((3 - x)/(3 + x))`
= `lim_(x -> 0) log((3 - x)/(3 + x))^(1/x)`
= `lim_(x -> 0) log((1 - x/3)/(1 + x/3))^(1/x)`
= `log[lim_(x -> 0) ((1 - x/3)^(1/x))/(1 + x/3)^(1/x)]`
= `log[({lim_(x -> 0)(1 - x/3)^((-3)/x)}^((-1)/3))/({lim_(x -> 0)(1 + x/3)^(3/x)}^(1/3))]`
= `log(("e"^(-1/3))/("e"^(1/3))) ...[because x -> 0"," ± x/3 ->0 and lim_(x->0)(1 + x)^(1/x) ="e"]`
= `log "e"^((-2)/3)`
= `-2/3*log "e"`
= `-2/3(1)`
= `-2/3`
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