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प्रश्न
Evaluate the following limit :
`lim_(x -> 0) [(log(3 - x) - log(3 + x))/x]`
उत्तर
`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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