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

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Answer 1 · 2016-06-22T08:54:18

1

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

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

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

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

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

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