How to find Maclaurin series of $y={(1+x)}^{1/x}$ at $x=0$ ?
I can't find it when it comes to $f'(0), f''(0)$. I have used limit, but somehow I don't know what to do with the $\frac{1}{x}$ term.
How to find Maclaurin series of $y={(1+x)}^{1/x}$ at $x=0$?
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$\begingroup$
calculus
taylor-expansion
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0Are you sure you want to be expanding this around $x = 0$? – 2017-01-25
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3Note that $$z=\frac1x\ln(1+x)-1=\frac1x\left(x-\frac12x^2+\frac13x^3+o(x^3)\right)-1=-\frac12x+\frac13x^2+o(x^2)$$ hence $$e^z=1+z+\frac12z^2+o(z^2)=1-\frac12x+\frac13x^2+\frac12\left(-\frac12x\right)^2+o(x^2)$$ that is $$y=(1+x)^{1/x}=e\cdot e^z=e-\frac{e}2x+\frac{11e}{24}x^2+o(x^2)$$ – 2017-01-25
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0@amcerbu Why not? – 2017-01-25
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0@Did: why post your answer as a comment? – 2017-01-25