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continuume
How do I use the hospital rule to solve this limit? lim x-> -infinity ( (2^x-2^(-x)) / 2^x+2^(-x) )
\[\frac{d}{dx} 2^x=\ln(2)2^x\] So I'm not sure how much it'll end up helping.
Your equation is\[\lim_{x \rightarrow -\infty}\frac{ 2^{x}-2^{-x} }{ 2^{x}+2^{-x} }\]? L'Hopital's Rule says that if \[\lim_{x \rightarrow c}f(x)=\lim_{x \rightarrow c}g(x)=0or \pm \infty\]and if\[\lim_{x \rightarrow c}\frac{ f'(x) }{ g'(x) }\]where g'(x) =/= 0 for all x in the domain (basically if the bottom isn't going to be 0) then\[\lim_{x \rightarrow c}\frac{ f(x) }{ g(x) }=\lim_{x \rightarrow c}\frac{ f'(x) }{ g'(x) }\]So the first thing you want to do is take \[\lim_{x \rightarrow -\infty}2 ^{x}-2^{-x}\]and\[\lim_{x \rightarrow -\infty}2^{x}+2^{-x}\]and see what they are. If they're both 0 or +- infinity, then you can use the rule. Take the derivative of the top and then take the derivative of the bottom, and then take the limit of the whole thing and see what you get. I'll stick around if you need help with any of that.
The biggest thing that'll help is that \[\ln 2 \times 2^{x}+\ln 2 \times2^{-x}=2 \times \ln 2 \times \cosh(\ln2 \times x)\]and\[\ln 2 \times 2^{x}-\ln 2 \times2^{-x}=2 \times \ln 2 \times \sinh(\ln2 \times x)\]
Thank you very much!