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 Calculus Calculus Math Forum

June 16th, 2014, 10:07 AM   #1
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Joined: Jun 2014
From: Canada

Posts: 1
Thanks: 0 Hard limit computation

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

Please help so lost!
Attached Images Screen Shot 2014-06-16 at 2.05.43 PM.png (8.9 KB, 18 views) June 16th, 2014, 11:35 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 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: 7,958 Thanks: 1146 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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Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus
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: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra For me $$\frac{d}{dx}(1+x)^\frac{1}{x} = -\frac{1}{x^2} \left( \log{(1+x)} - \frac{x}{x+1} \right)(1+x)^\frac{1}{x}$$ Tags computation, hard, limit Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ZardoZ Complex Analysis 0 May 4th, 2013 01:12 PM unm Number Theory 1 November 17th, 2012 09:58 AM mathbalarka Calculus 9 August 31st, 2012 05:40 AM stuart clark Calculus 1 March 14th, 2011 09:33 PM 4011 Calculus 12 May 7th, 2007 04:23 PM

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