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kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow +\infty} (1+ \frac{k}{x})^{x}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
any guesses?
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
it's in indeterminate form \[1^{\infty}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
since on definition of \(e\) is \(\lim_{x\to \infty}(1+\frac{1}{x})^x\) it will be no surprise that \[\lim_{x\to \infty}\left(1+\frac{k}{x}\right)^x=e^k\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
if that is an unacceptable answer, take the log, take the limit, which will be \(k\) and then exponentiate to get \(e^k\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
start 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
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
let \[y=(1+\frac{k}{x})^{x}\] \[\ln{y} =x\ln(1+\frac{k}{x})\] then, \[\lim_{x \rightarrow +\infty} x \ln (1+\frac{k}{x})\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
yup then go to \[\frac{\ln(1+\frac{k}{x})}{\frac{1}{x}}\] and you should be good to go
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
=0 *(+Infinity)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
right rewrite as i did above to get \(\frac{0}{0}\)
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
we can rewrite it to this form \[\lim_{x \rightarrow +\infty} \frac{\ln(1+ \frac{k}{x})}{\frac{1}{x}}\]
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
=\[\frac{0}{0}\]
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
should we use the L' Hospital Rule now?
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
what's the derivative of \[\ln (1+\frac{k}{x})?\] is it \[\frac{x^{2}}{(x+k)^{2}}?\]
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
ok so it's \[\frac{k}{x^{2}+kx}\]
 one year ago

sunnymony Group TitleBest ResponseYou've already chosen the best response.0
u know 1^infinite will be =o
 one year ago

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
i don't think it is.
 one year ago
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