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June 16th, 2014, 10:07 AM   #1
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Hard limit computation

I need to do this using L'hospitals or log differentiation

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 June 16th, 2014, 11:35 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra I think all derivatives of the numerator grow without bound as $x \to 0$.
 June 16th, 2014, 01:43 PM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus $$\lim_{x\to 0}\frac{\left(x+1\right)^{\frac{1}{x}}-\mathbb{e}}{x}=\lim_{x\to 0 }\frac{\mathbb{e}-\mathbb{e}\frac{x}{2}+\mathcal{O}(x^2)-\mathbb{e}}{x}=\lim_{x\to 0 }-\frac{\mathbb{e}}{2}+\mathcal{O}(x)=-\frac{\mathbb{e}}{2}$$
 June 16th, 2014, 01:56 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond Using L'Hospital's rule several times, $\displaystyle \frac{d}{dx}(x+1)^{1/x}=-\frac1x\left(\ln(x+1)^{1/x}-\frac{1}{x+1}\right)(x+1)^{1/x}$ $\displaystyle \Rightarrow-\frac{\ln(x+1)^{1/x}-\frac{1}{x+1}}{x}\cdot e$ $\displaystyle \Rightarrow-\left(\frac{\frac{x}{x+1}-\ln(x+1)}{x^2}+\frac{1}{(x+1)^2}\right)\cdot e$ $\displaystyle \Rightarrow-\left(\frac{\frac{x}{x+1}-\ln(x+1)}{x^2}+1\right)\cdot e$ $\displaystyle \Rightarrow-\left(\frac{\frac{1}{(x+1)^2}-\frac{1}{x+1}}{2x}+1\right)\cdot e$ $\displaystyle \Rightarrow-\left(\frac{\frac{-2}{(x+1)^3}+\frac{1}{(x+1)^2}}{2}+1\right)\cdot e$ $\displaystyle \Rightarrow\lim_{x\to0}\frac{(x+1)^{1/x}-e}{x}=-\frac{e}{2}$ Thanks from ZardoZ
June 16th, 2014, 02:05 PM   #5
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Quote:
 Originally Posted by greg1313 Using L'Hospital's rule several times,
Nice!

 June 16th, 2014, 02:37 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra For me $$\frac{d}{dx}(1+x)^\frac{1}{x} = -\frac{1}{x^2} \left( \log{(1+x)} - \frac{x}{x+1} \right)(1+x)^\frac{1}{x}$$

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