Question

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

Answer 1

`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`