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continuume
Group Title
How do I use the hospital rule to solve this limit?
lim x> infinity ( (2^x2^(x)) / 2^x+2^(x) )
 one year ago
 one year ago
continuume Group Title
How do I use the hospital rule to solve this limit? lim x> infinity ( (2^x2^(x)) / 2^x+2^(x) )
 one year ago
 one year ago

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jishan Group TitleBest ResponseYou've already chosen the best response.0
dw:1354999747553:dw
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{d}{dx} 2^x=\ln(2)2^x\] So I'm not sure how much it'll end up helping.
 one year ago

Aylin Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

Aylin Group TitleBest ResponseYou've already chosen the best response.1
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\[\ln 2 \times 2^{x}\ln 2 \times2^{x}=2 \times \ln 2 \times \sinh(\ln2 \times x)\]
 one year ago

continuume Group TitleBest ResponseYou've already chosen the best response.0
Thank you very much!
 one year ago

Aylin Group TitleBest ResponseYou've already chosen the best response.1
You're welcome! :D
 one year ago
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