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kaiz122
 3 years ago
help me here. please
kaiz122
 3 years ago
help me here. please

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow +\infty} (1+ \frac{k}{x})^{x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's in indeterminate form \[1^{\infty}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since on definition of \(e\) is \(\lim_{x\to \infty}(1+\frac{1}{x})^x\) it will be no surprise that \[\lim_{x\to \infty}\left(1+\frac{k}{x}\right)^x=e^k\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if that is an unacceptable answer, take the log, take the limit, which will be \(k\) and then exponentiate to get \(e^k\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0start with \[x\ln(1+\frac{1}{x})\] which is now \(\infty\times 0\) then the usual trick is to rewrite as \[\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}\] so it is now \(\frac{0}{0}\) and then use l'hopital

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let \[y=(1+\frac{k}{x})^{x}\] \[\ln{y} =x\ln(1+\frac{k}{x})\] then, \[\lim_{x \rightarrow +\infty} x \ln (1+\frac{k}{x})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup then go to \[\frac{\ln(1+\frac{k}{x})}{\frac{1}{x}}\] and you should be good to go

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right rewrite as i did above to get \(\frac{0}{0}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we can rewrite it to this form \[\lim_{x \rightarrow +\infty} \frac{\ln(1+ \frac{k}{x})}{\frac{1}{x}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0should we use the L' Hospital Rule now?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what's the derivative of \[\ln (1+\frac{k}{x})?\] is it \[\frac{x^{2}}{(x+k)^{2}}?\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so it's \[\frac{k}{x^{2}+kx}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u know 1^infinite will be =o
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