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 June 14th, 2014, 11:43 PM #1 Newbie   Joined: Jun 2014 From: chennai Posts: 17 Thanks: 0 new to limits help! IMG_20140706_135215.jpg can I do the encircled step.the final answer is correct.but I dont know if the step is help pls . June 15th, 2014, 06:00 AM #2 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 With Del Hospital's Rule ..... $$\hspace{100pt}\boxed{\begin{eqnarray} \lim_{x\to\frac{\pi}{4}}(2-\tan(x))^{\frac{1}{\ln(\tan(x))}}&=&\lim_{x\to\fra c{\pi}{4}}\exp\left(\frac{\ln\left(2-\tan(x)\right)}{\ln(\tan(x))}\right) \\ &=&\exp\left(\lim_{x\to\frac {\pi}{4}}\frac{\ln(2-\tan(x)}{\ln(\tan(x))}\right)\\&=&\exp\left(\lim_{ x\to\frac {\pi}{4}}\frac{\ln\left(2-\frac{\sin(x)}{\cos(x)}\right)}{\ln\left(\frac{\si n(x)}{\cos(x)}\right)}\right)\\&\overset{DLH}{=}&\ exp\left(\lim_{x\to\frac{\pi}{4}}\frac{\sin(x)}{si n(x)-2\cos(x)}\right)\\&=&\exp\left(\frac{\frac{\sqrt{2 }}{2}}{\frac{\sqrt{2}}{2}-\sqrt{2}}\right)\\&=&\mathbb{e}^{-1}\end{eqnarray}}$$ Last edited by ZardoZ; June 15th, 2014 at 06:07 AM. June 15th, 2014, 06:48 AM   #3
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Quote:
 Originally Posted by ZardoZ $$\hspace{100pt}\boxed{\begin{eqnarray} \lim_{x\to\frac{\pi}{4}}(2-\tan(x))^{\frac{1}{\ln(\tan(x))}}&=&\lim_{x\to\fra c{\pi}{4}}\exp\left(\frac{\ln\left(2-\tan(x)\right)}{\ln(\tan(x))}\right) \\ &=&\exp\left(\lim_{x\to\frac {\pi}{4}}\frac{\ln(2-\tan(x)}{\ln(\tan(x))}\right)\\&=&\exp\left(\lim_{ x\to\frac {\pi}{4}}\frac{\ln\left(2-\frac{\sin(x)}{\cos(x)}\right)}{\ln\left(\frac{\si n(x)}{\cos(x)}\right)}\right)\\&\overset{DLH}{=}&\ exp\left(\lim_{x\to\frac{\pi}{4}}\frac{\sin(x)}{si n(x)-2\cos(x)}\right)\\&=&\exp\left(\frac{\frac{\sqrt{2 }}{2}}{\frac{\sqrt{2}}{2}-\sqrt{2}}\right)\\&=&\mathbb{e}^{-1}\end{eqnarray}}$$
i know this method. i am not asking how you do it.i want to know if the step encircled is correct or not.thnk u for ur time June 15th, 2014, 07:16 AM #4 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 Without knowing that the limit exists, why do you denote it by an 'L'. I always avoid this. June 15th, 2014, 08:33 PM   #5
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 Originally Posted by ZardoZ Without knowing that the limit exists, why do you denote it by an 'L'. I always avoid this.
i am assuming the limit to be some L.then applying ln on both sides. June 15th, 2014, 09:40 PM   #6
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Hello, harryjobs!

Quote:
 $\displaystyle\lim_{x\to\frac{\pi}{4}} (2-\tan x)^{\frac{1}{\ln(\tan x)}}$
Let $y \:=\:(2-\tan x)^{\frac{1}{\ln(\tan x)}}$

Take logs: $\:\ln y \;=\;\ln(2-\tan x)^{\frac{1}{\ln(\tan x)}} \;=\;\dfrac{\ln(2-\tan x)}{\ln(\tan x)} \;\;\to\;\;\frac{0}{0}$

Apply L'Hopital: $\:\ln y \;=\;\dfrac{\dfrac{-\sec^2\!x}{2-\tan x}}{\dfrac{\sec^2\!x}{\tan x}} \;=\;\dfrac{-\tan x}{2-\tan x}$

$\displaystyle \lim_{x\to\frac{\pi}{4}} (\ln y) \;=\;\lim_{x\to\frac{\pi}{4}} \dfrac{-\tan x}{2-\tan x} \;=\;\frac{-1}{2-1} \;=\;-1$

Hence: $\:\ln y \:=\:-1 \quad\Rightarrow\quad y \:=\:e^{-1}$

Therefore: $\:y \:=\:\dfrac{1}{e}$ June 15th, 2014, 10:59 PM   #7
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Quote:
 Originally Posted by harryjobs can I do the encircled step
Yes. While the $\lim$ symbol is there, $y$ is non-zero, so the numerator and denominator in the circled quotient are both non-zero. Tags limits 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 velu1996 Calculus 6 January 3rd, 2013 02:28 PM kadmany Calculus 9 March 18th, 2011 07:16 AM lilwayne Calculus 18 September 23rd, 2010 03:39 PM yehoram Calculus 8 June 25th, 2010 09:40 AM mathman2 Calculus 11 October 20th, 2009 07:08 PM

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