Question

Evaluate the following limit : 

`lim_(x -> 0) [(8^sinx - 2^tanx)/("e"^(2x) - 1)]`

Answer 1

`lim_(x -> 0) (8^sinx - 2^tanx)/("e"^(2x) - 1)`

= `lim_(x -> 0) ((8^sinx - 1)(2^tanx - 1))/("e"^(2x) - 1)`

= `lim_(x -> 0) (((8^sinx - 1) - (2^tanx - 1))/x)/(("e"^(2x) - 1)/x)   ...[("Divide numerator and"),("Denominator by"  x.),(because x -> 0 therefore x ≠ 0)]`

= `(lim_(x -> 0) ((8^sinx - 1)/x - (2^tanx - 1)/x))/(lim_(x -> 0) ("e"^(2x) - 1)/x)`

= `(lim_(x -> 0) ((8^sinx - 1)/sinx* sinx/x - (2^tanx - 1)/tanx * tanx/x))/(lim_(x -> 0) (e^(2x) - 1)/x)`

= `((lim_(x -> 0) (8^sinx - 1)/sinx)(lim_(x -> 0) sinx/x) - (lim_(x -> 0) (2^tanx - 1)/tanx)(lim_(x -> 0) tanx/x))/((lim_(x -> 0) ("e"^(2x) - 1)/(2x)) xx 2)`

= `(log8(1) - (log2)(1))/((1) xx 2)  ...[(because x -> 0","  2x -> 0","),(sin x -> 0"," tanx -> 0),(lim_(x -> 0) ("a"^x - 1)/x = log"a")]`

= `(log  8/2)/2`

= `(log4)/2`

= `(log(2)^2)/2`

= `(2log2)/2`

= log 2