kaiz122
help me here. please
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kaiz122
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\[\lim_{x \rightarrow +\infty} (1+ \frac{k}{x})^{x}\]
anonymous
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any guesses?
kaiz122
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it's in indeterminate form \[1^{\infty}\]
anonymous
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since on definition of \(e\) is \(\lim_{x\to \infty}(1+\frac{1}{x})^x\) it will be no surprise that
\[\lim_{x\to \infty}\left(1+\frac{k}{x}\right)^x=e^k\]
anonymous
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if that is an unacceptable answer, take the log, take the limit, which will be \(k\) and then exponentiate to get \(e^k\)
anonymous
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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
kaiz122
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let \[y=(1+\frac{k}{x})^{x}\]
\[\ln{y} =x\ln(1+\frac{k}{x})\] then,
\[\lim_{x \rightarrow +\infty} x \ln (1+\frac{k}{x})\]
anonymous
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yup then go to
\[\frac{\ln(1+\frac{k}{x})}{\frac{1}{x}}\] and you should be good to go
kaiz122
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=0 *(+Infinity)
anonymous
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right
rewrite as i did above to get \(\frac{0}{0}\)
kaiz122
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we can rewrite it to this form \[\lim_{x \rightarrow +\infty} \frac{\ln(1+ \frac{k}{x})}{\frac{1}{x}}\]
kaiz122
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=\[\frac{0}{0}\]
kaiz122
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should we use the L' Hospital Rule now?
anonymous
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yes
kaiz122
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what's the derivative of \[\ln (1+\frac{k}{x})?\]
is it \[\frac{x^{2}}{(x+k)^{2}}?\]
phi
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no.
kaiz122
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ok so it's \[\frac{-k}{x^{2}+kx}\]
sunnymony
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u know 1^infinite will be =o
kaiz122
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i don't think it is.