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June 14th, 2014, 11:43 PM   #1
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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 .
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June 15th, 2014, 06:00 AM   #2
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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.
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June 15th, 2014, 06:48 AM   #3
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Quote:
Originally Posted by ZardoZ View Post
$$\hspace{100pt}\boxed{\begin{eqnarray} \lim_{x\to\frac{\pi}{4}}(2-\tan(x))^{\frac{1}{\ln(\tan(x))}}&=&\lim_{x\to\fra c{\pi}{4}}\exp\left(\frac{\ln\left(2-\tan(x)\right)}{\ln(\tan(x))}\right) \\ &=&\exp\left(\lim_{x\to\frac {\pi}{4}}\frac{\ln(2-\tan(x)}{\ln(\tan(x))}\right)\\&=&\exp\left(\lim_{ x\to\frac {\pi}{4}}\frac{\ln\left(2-\frac{\sin(x)}{\cos(x)}\right)}{\ln\left(\frac{\si n(x)}{\cos(x)}\right)}\right)\\&\overset{DLH}{=}&\ exp\left(\lim_{x\to\frac{\pi}{4}}\frac{\sin(x)}{si n(x)-2\cos(x)}\right)\\&=&\exp\left(\frac{\frac{\sqrt{2 }}{2}}{\frac{\sqrt{2}}{2}-\sqrt{2}}\right)\\&=&\mathbb{e}^{-1}\end{eqnarray}}$$
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
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June 15th, 2014, 07:16 AM   #4
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Without knowing that the limit exists, why do you denote it by an 'L'. I always avoid this.
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June 15th, 2014, 08:33 PM   #5
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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.
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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: $\:y \:=\:\dfrac{1}{e}$
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June 15th, 2014, 10:59 PM   #7
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Quote:
Originally Posted by harryjobs View Post
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.
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