My Math Forum limit help required

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 July 8th, 2013, 09:21 AM #1 Newbie   Joined: Jul 2013 Posts: 1 Thanks: 0 limit help required Can anyone help me with the proof of the following limit please?? limx?1 (C^(1-x)-1) / (1-x) = lnC Thank you in advance.
 July 8th, 2013, 01:26 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond Re: limit help required Hi, Levon and welcome to the forum. When you have a question, please post it in its own topic. Thanks. One way to do this problem is with L'Hospital's Rule: $\lim_{x\to a}\,\frac{f(x)}{g(x)}\,=\,\lim_{x\to a}\,\frac{f'(x)}{g#39;(x)}$ if $\lim_{x\to a}f(x)\,=\,\lim_{x\to a}g(x)\,=\,0.$ A complete statement of the rule is given here. A solution with L'Hospital's Rule: It should be clear that the derivative of the function in the denominator is -1. $\frac{d}{dx}\,C^{1-x}\,=\,-\ln(C)\,\cdot\,C^{1-x}$ calculated as follows, $y\,=\,C^{1-x} \\ \ln(y)\,=\,(1-x)\ln(C) \\ \frac{y'}{y}\,=\,-\ln(C) \\ y'\,=\,-\ln(C)\,\cdot\,C^{1-x}$ so we have $\lim_{x\to1}\,\frac{C^{1-x}\,-\,1}{1\,-\,x}\,=\,\frac{-\ln(C)\,\cdot\,C^{1-1}}{-1}\,=\,\ln(C)$
 July 9th, 2013, 01:06 PM #3 Senior Member   Joined: Aug 2012 From: New Delhi, India Posts: 157 Thanks: 0 Re: limit help required Aliter $\text{Let }1-x=t \text{As }x \rightarrow 1\text{ } t \rightarrow 0 L=\lim_{t \to 0} \frac{C^t -1}{t} = ln C$

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