the questions in the attached file...

thanks.

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- Dec 14th 2006, 01:54 AMPhoebe83I need some helpthe questions in the attached file...

thanks. - Dec 14th 2006, 04:07 AMtopsquark
$\displaystyle \lim_{x \to 0} \frac{sin(x) - tan^{-1}(x)}{x^2 ln(1 + x)}$

This is of the form 0/0 and so can be done by using L'Hopital's rule. However taking the derivative of this is going to be long and susceptible to errors. What I would suggest is that we replace the sine, inverse tangent, and natural log functions with their Taylor expansions for x near 0:

$\displaystyle sin(x) \approx x - \frac{x^3}{6}$

$\displaystyle tan^{-1}(x) \approx x - \frac{x^3}{3}$

$\displaystyle ln(1 + x) \approx x - \frac{x^2}{2}$

So

$\displaystyle \lim_{x \to 0} \frac{sin(x) - tan^{-1}(x)}{x^2 ln(1 + x)} \approx \frac{x - \frac{x^3}{6} - x + \frac{x^3}{3}}{x^2 \left (x - \frac{x^2}{2} \right ) }$

= $\displaystyle \lim_{x \to 0} \frac{ \frac{x^3}{6} }{x^3 - \frac{x^4}{2} }$

= $\displaystyle \lim_{x \to 0} \frac{x^3}{6x^3 - 3x^4}$

= $\displaystyle \lim_{x \to 0} \frac{1}{6 - 3x}$

= $\displaystyle \frac{1}{6}$

-Dan - Dec 14th 2006, 04:27 AMtopsquark
$\displaystyle \lim_{x \to 0} \left ( \frac{1}{sin^2(x)} - \frac{1}{x^2} \right )$

Again, I would employ the Taylor series trick here (because I just tried the L'Hopital's route and the 0/0 didn't go away after two applications and things were again getting ugly.)

First combine the fractions:

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

$\displaystyle sin^2(x) \approx x^2 - \frac{x^4}{3}$

$\displaystyle \lim_{x \to 0} \frac{x^2 - sin^2(x)}{x^2sin^2(x)} = \lim_{x \to 0} \frac{x^2 - x^2 + \frac{x^4}{3}}{x^2 \left (x^2 - \frac{x^4}{3} \right ) }$

= $\displaystyle \lim_{x \to 0} \frac{\frac{x^4}{3}}{x^4 - \frac{x^6}{3}}$

= $\displaystyle \lim_{x \to 0} \frac{\frac{1}{3}}{1 - \frac{x^2}{3}}$

= $\displaystyle \frac{1}{3}$

-Dan - Dec 14th 2006, 04:45 AMPhoebe83Thanks!! one question
I loved the Tylor trick...

Can you explain why we can take the Taylor expansions for x near 0? is it because the Lim is for x near 0? - Dec 14th 2006, 04:58 AMtopsquark
Yes, that's exactly it. Normally I would consider such an approach "cheating" but it looks like the problems you were given were designed to make an application of L'Hopital's rule difficult if not impossible. And, as I'm a Physicist and not a Mathematician I have no problems with "cheating" if it gets me the answers. :)

-Dan