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

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

kaiz122 Group TitleBest ResponseYou've already chosen the best response.0
it's in indeterminate form \[1^{\infty}\]
 2 years 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 \[\lim_{x\to \infty}\left(1+\frac{k}{x}\right)^x=e^k\]
 2 years 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 \(e^k\)
 2 years 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
 2 years 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, \[\lim_{x \rightarrow +\infty} x \ln (1+\frac{k}{x})\]
 2 years 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
 2 years ago

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

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

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

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

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

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

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

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

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