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November 15th, 2016, 08:40 AM  #11 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  
November 15th, 2016, 09:37 AM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
That is incorrect. The limit is 1. $$\begin{align*} f(x) &= \frac1x + e^x = \frac{1+xe^{x}}{x} \\ g(x) = \frac1{f(x)} &= \frac{x}{1+xe^{x}} \\ &\approx x\big(1x(1x)\big) = xx^2+x^3 \\ g'(x) & \approx 1  2x \\ \lim_{x \to 0} g'(x) = 1 \end{align*}$$ 
November 15th, 2016, 11:03 AM  #13 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
Thanks. I made an algebraic mistake. Suppose: f(x)=1/x+g(x) Then $\displaystyle \lim_{x\rightarrow 0}\frac{\mathrm{d} \frac{1}{f(x))}}{\mathrm{d} x}= \lim_{x\rightarrow 0}\frac{1x^{2}g'(x)}{(1+xg(x))^{2}}$ Which depends on behavior of g(x) From https://en.wikipedia.org/wiki/Riemann_zeta_function $\displaystyle \zeta (s)={\frac {1}{s1}}+\sum _{n=0}^{\infty }{\frac {(1)^{n}}{n!}}\gamma _{n}\;(s1)^{n}$.  $\displaystyle \zeta$ (s) = $\displaystyle {\frac {1}{s1}}$+g(s1) In this case Lim is indeed 1. skipjack is right. I still question the argument that if f(x) $\displaystyle \approx$ h(x). f'(x) $\displaystyle \approx$ h'(x) Last edited by zylo; November 15th, 2016 at 11:05 AM. 
November 15th, 2016, 02:49 PM  #14 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  Notice that the approach I took demonstrates that the derivative is 1 for every $g(x)$ having a Maclaurin expansion. Also (I think) for every $g(x)$ that is continuous at zero. Again, if you consider series expansions, the result is pretty obvious, at least in the case that $h(x)$ is a polynomial. 
November 15th, 2016, 06:19 PM  #15  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Quote:
Replace x with s complex to get $\displaystyle \zeta$(s) Wiki article says $\displaystyle \zeta$(s) is analytic and therefore can be expressed as a Laurent series. $\displaystyle \zeta$(s)=$\displaystyle \sum a_{n}(s1)^{n}$ Therefore $\displaystyle \zeta$(x) = $\displaystyle \sum a_{n}(x1)^{n}$ and above quote works. The point is that the analyticity of $\displaystyle \zeta$(s) allows you to find a power series for $\displaystyle \zeta$(x). Last edited by skipjack; November 16th, 2016 at 12:04 AM.  

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derivative, function, infinity, involving, limit, zeta, zeta function 
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