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

1 Answer

Jun 22, 2016

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#

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