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\[\lim_{x \rightarrow +\infty} (1+ \frac{k}{x})^{x}\]
any guesses?
it's in indeterminate form \[1^{\infty}\]

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Other answers:

since on definition of \(e\) is \(\lim_{x\to \infty}(1+\frac{1}{x})^x\) it will be no surprise that \[\lim_{x\to \infty}\left(1+\frac{k}{x}\right)^x=e^k\]
if that is an unacceptable answer, take the log, take the limit, which will be \(k\) and then exponentiate to get \(e^k\)
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
let \[y=(1+\frac{k}{x})^{x}\] \[\ln{y} =x\ln(1+\frac{k}{x})\] then, \[\lim_{x \rightarrow +\infty} x \ln (1+\frac{k}{x})\]
yup then go to \[\frac{\ln(1+\frac{k}{x})}{\frac{1}{x}}\] and you should be good to go
=0 *(+Infinity)
right rewrite as i did above to get \(\frac{0}{0}\)
we can rewrite it to this form \[\lim_{x \rightarrow +\infty} \frac{\ln(1+ \frac{k}{x})}{\frac{1}{x}}\]
should we use the L' Hospital Rule now?
what's the derivative of \[\ln (1+\frac{k}{x})?\] is it \[\frac{x^{2}}{(x+k)^{2}}?\]
  • phi
ok so it's \[\frac{-k}{x^{2}+kx}\]
u know 1^infinite will be =o
i don't think it is.

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