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claudiag Group Title

What is the limit as x goes to infinity of (1+2/x)^x and how do I get it?

  • 3 years ago
  • 3 years ago

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  1. dumbcow Group Title
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    put in indeterminate form then apply L'Hopitals rule by differentiating top and bottom rewrite it as \[\frac{(x+2)^x}{x^x}\]

    • 3 years ago
  2. claudiag Group Title
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    I 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?

    • 3 years ago
  3. claudiag Group Title
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    through* sorry

    • 3 years ago
  4. dumbcow Group Title
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    \[y = x^{x} \rightarrow \ln y= x \ln x\] Now differentiate both sides \[\frac{1}{y}\frac{dy}{dx} = \ln x +1\] multiply by y on both sides where y = x^x \[\frac{dy}{dx} = x^{x}(\ln x+1)\]

    • 3 years ago
  5. dumbcow Group Title
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    ok 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 inf-inf/0 which is indeterminate so apply L'hopitals rule and differentiate top and bottom \[e ^{\lim_{x \rightarrow \infty}\frac{\frac{1}{x+2}-\frac{1}{x}}{-\frac{1}{x ^{2}}}}\]

    • 3 years ago
  6. dumbcow Group Title
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    combine 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

    • 3 years ago
  7. claudiag Group Title
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    Thank'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!!

    • 3 years ago
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