Question

Using L’Hospital Rule, evaluate: `lim_(x->0)  (8^x - 4^x)/(4x
)`

Answer 1

`lim_(x-.0) (8^x-4^x)`
`lim_(x->0)(8^xlog8 - 4^x log 4)`

`= (8^0 log 8 - 4^0 log 4)/4`
`=1/4log2`