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dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1put in indeterminate form then apply L'Hopitals rule by differentiating top and bottom rewrite it as \[\frac{(x+2)^x}{x^x}\]

claudiag
 3 years ago
Best ResponseYou've already chosen the best response.0I think I get it but, the derivation is kind of weird... or I don't really know if I'm doing it right... would it bother you too much to walk me throught it?

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1\[y = x^{x} \rightarrow \ln y= x \ln x\] Now differentiate both sides \[\frac{1}{y}\frac{dy}{dx} = \ln x +1\] multiply by y on both sides where y = x^x \[\frac{dy}{dx} = x^{x}(\ln x+1)\]

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1ok i realized i was wrong in my approach, those derivatives do get messy and dont help find a solution. my apologies, here is the correct solution step by step transform problem using natural log identity, e^ln(x) = x \[e ^{\lim_{x \rightarrow \infty}\ln (\frac{x+2}{x})^{x}}\] use log rules to expand and move x to denominator \[e ^{\lim_{x \rightarrow \infty}\frac{\ln (x+2) \ln x}{\frac{1}{x}}}\] here we notice if we evaluate limit we get infinf/0 which is indeterminate so apply L'hopitals rule and differentiate top and bottom \[e ^{\lim_{x \rightarrow \infty}\frac{\frac{1}{x+2}\frac{1}{x}}{\frac{1}{x ^{2}}}}\]

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1combine fractions on top then multiply by x^2 on top and bottom \[e ^{\lim_{x \rightarrow \infty}\frac{2x}{x+2}}\] if you evaluate again you'll get infinity over infinity so differentiate again \[e ^{\lim_{x \rightarrow \infty}2}\rightarrow e ^{2}\] so the limit of the function is e^2

claudiag
 3 years ago
Best ResponseYou've already chosen the best response.0Thank's I'll try it out. I don't really understand the first step in which you transform the problem using the idientity, I don't understand how is that possible. But I do understand the resto of it so I'll find my teacher and ask him about that. Thanks a lot!!
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