Question

How do you find the limit of `(2^x-3^-x)/(2^x+3^-x)` as x approaches infinity?

Answer 1

1

Explanation:

`L = lim_{x \to \infty} (2^x-3^-x)/(2^x+3^-x)`

`= lim_{x \to \infty} (2^x-3^-x)/(2^x+3^-x) * (2^{-x})/(2^{-x})`

`= lim_{x \to \infty} (1-3^{-x}*2^{-x})/(1+3^{-x}*2^{-x})`

`= lim_{x \to \infty} (1 - 6^{-x} )/(1+6^{-x})`

`lim_{x \to infty} 6^{-x} = 0 \implies L = 1`