Question

How do you evaluate `(2^x-3^-x)/(2^x+3^-x)` as x approaches infinity?

Answer 1

`lim_{x->oo}(2^x-3^-x)/(2^x+3^-x) =1`

Explanation:

`(2^x-3^-x)/(2^x+3^-x) =(1-6^{-x})/(1+6^{-x})`

`lim_{x->oo}(2^x-3^-x)/(2^x+3^-x) = (1-lim_{x->oo}6^{-x})/(1+lim_{x->oo}6^{-x}) = 1`