limit

**dhiab** · July 9th 2011, 06:29 AM

Calculat :

**FernandoRevilla** · July 9th 2011, 06:48 AM

Use $L=\lim_{x\to a}f(x)^{g(x)}=e^{\lambda}$ where $\lambda=\lim_{x\to a}\log f(x)^{g(x)}$ .

**Sudharaka** · July 9th 2011, 06:48 AM

Originally Posted by dhiab

Calculat :

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Dear dhiab,

$\lim_{x\rightarrow 0}\left(\frac{2-\cos x}{1-\cos x}\right)^{\sin^{2}x}$

$=\lim_{x\rightarrow 0}e^{\left(\sin^{2}x\right)\ln\left(\frac{2-\cos x}{1-\cos x}\right)}$

$=exp\left(\lim_{x\rightarrow 0}\frac{\ln\left(\frac{2-\cos x}{1-\cos x}\right)}{\frac{1}{\sin^{2}x}}\right)$

Use the L'Hospital's rule to evaluate the limit. Hope you can do it from here.

**Archie Meade** · July 9th 2011, 07:26 AM

Originally Posted by dhiab

Calculat :

You could also use

$\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x=e$

while

$\lim_{x \to 0^+}\left(1+\frac{1}{x}\right)^x=1$

$\lim_{x \to 0}\left(\frac{2-cosx}{1-cosx}\right)^{sin^2x}=\lim_{x \to 0}\left(\frac{1+1-cosx}{1-cosx}\right)^{1-cos^2x}$

$=\lim_{x \to 0}\left[\left[\left(1+\frac{1}{1-cosx}\right)^{1-cosx}\right]^{1+cosx}\right]$

$=\lim_{x \to 0}\left[\left[\lim_{y \to 0}\left(1+\frac{1}{y}\right)^{y}\right]^{1+cosx}\right]$

EDIT:

A glaring error in the limit definition for "e"
and some sloppiness in further lines have been edited.
Thanks to CaptainBlack and Opalg.

**CaptainBlack** · July 9th 2011, 08:08 AM

Originally Posted by Archie Meade

You could also use

$e^x=\lim_{x \to 0}\left(1+\frac{1}{x}\right)^x$

Hence

$\lim_{x \to 0}\left(\frac{2-\cos x}{1-\cos x}\right)^{\sin^2x}=\lim_{x \to 0}\left(\frac{1+1-\cos x}{1-\cos x}\right)^{1-\cos^2x}$

$\color{red}=\left[\lim_{x \to 0}\left(1+\frac{1}{1-\cos x}\right)^{1-\cos x}\right]^{1+\cos x}$

$=\lim_{x \to 0}\left[e^{1-\cos x}\right]^{1+\cos x}$

$=\lim_{x \to 0}e^{1-\cos^2x}=\lim_{x \to 0}e^{\sin^2x}$

You need to be more careful about line 5 (and onwards).

CB

**Opalg** · July 9th 2011, 08:39 AM

Originally Posted by CaptainBlack

Originally Posted by Archie Meade

You could also use

$\color{red}e^x=\lim_{x \to 0}\left(1+\frac{1}{x}\right)^x$

Hence

$\lim_{x \to 0}\left(\frac{2-cosx}{1-cosx}\right)^{sin^2x}=\lim_{x \to 0}\left(\frac{1+1-cosx}{1-cosx}\right)^{1-cos^2x}$

$=\left[\lim_{x \to 0}\left(1+\frac{1}{1-cosx}\right)^{1-cosx}\right]^{1+cosx}$

$=\lim_{x \to 0}\left[e^{1-cosx}\right]^{1+cosx}$

$=\lim_{x \to 0}e^{1-cos^2x}=\lim_{x \to 0}e^{sin^2x}$

You need to be more careful about line 5 (and onwards).

CB

... actually needs to be more careful from line 1 onwards!

**CaptainBlack** · July 9th 2011, 09:28 AM

Originally Posted by Archie Meade

You could also use

$e^x=\lim_{x \to 0}\left(1+\frac{1}{x}\right)^x$

Hence

$\lim_{x \to 0}\left(\frac{2-cosx}{1-cosx}\right)^{sin^2x}=\lim_{x \to 0}\left(\frac{1+1-cosx}{1-cosx}\right)^{1-cos^2x}$

$=\left[\lim_{x \to 0}\left(1+\frac{1}{1-cosx}\right)^{1-cosx}\right]^{1+cosx}$

$=\lim_{x \to 0}\left[e^{1-cosx}\right]^{1+cosx}$

$=\lim_{x \to 0}e^{1-cos^2x}=\lim_{x \to 0}e^{sin^2x}$

OK what can we salvage from this?

As (you may feel the need to prove this if you don't know it already):

$\lim_{x\to 0}\left(1+\frac{1}{x}\right)^x=1$

We may argue:

$\begin{array}{lcl}\displaystyle \lim_{x \to 0}\left(\frac{2-\cos(x)}{1-\cos(x)}\right)^{\sin^2(x)}&=&\displaystyle\lim_{x \to 0}\left(\frac{1+1-\cos(x)}{1-\cos(x)}\right)^{1-\cos^2(x)}\\ \\ &=&\displaystyle \lim_{x \to 0}\left\{ \left[ \left(1+\frac{1}{1-\cos(x)}\right)^{1-\cos(x)}\right]^{1+\cos(x)} \right\}\\ \\&=&\displaystyle \left\{\lim_{x \to 0} \left[ \left(1+\frac{1}{1-\cos(x)}\right)^{1-\cos(x)}\right]\right\}^{\displaystyle\lim_{x\to 0}(1+\cos(x))}\\ \\&=&1^2=1 \end{array}$

Though a little care is required to justify the second from last line above (as always, it does help to know what the limit should be).

CB

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