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lim as x approaches positive infinity of (1 + 1/x) ^ x
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I would use the fact that e^(ln(y))=y
\[\lim_{x \rightarrow \infty}e^{x \ln(1+\frac{1}{x})}=\lim_{x \rightarrow \infty} e^{\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}}\]
use l'hosptal rule now on that exponent
lim as x approaches infinity of e ^ {(ln(1 + 1/x) over 1/x} but i am stumped after that, it must be super obvious but i dont get it
first of all the exponent gives us 0/0 right so we can use l'hosptal
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what is the derivative of ln(1+1/x) and the derivative of 1/x?
the derivative would be -1/x^3 and -1/x^2
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