A Limit (Maybe using the infinite series?)

**Ted** · June 24th 2010, 03:48 AM

Hello.
I stopped at this one:

$\lim_{ x \to 0} \dfrac{1-\frac{1}{2}x^2-cos\left( \frac{x}{1-x^2}\right)}{x^4}$

I did not try the L'Hospital's Rule, Am searching for a perfect solution.
Maybe using the infinite series ?
But it seems brutal.

Any help?

**Utherr** · June 24th 2010, 04:08 AM

transform cos into sin using $\cos{2x}=1-2\sin^2{x}$ then u can apply (sin x)/x -> 1 limit... then see what you have left

**simplependulum** · June 24th 2010, 04:08 AM

$\lim_{x\to 0 } \frac{ 1 - \frac{x^2}{2} - \cos( \frac{x}{1-x^2} ) }{x^4}$

$= - \frac{1}{24} + \lim_{x\to 0 } \frac{ 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cos( \frac{x}{1-x^2} ) }{x^4 }$

$= - \frac{1}{24} + \lim_{x\to 0 } \frac{ \cos(x) + O(x^6) - \cos( \frac{x}{1-x^2} ) }{x^4 }$

$= - \frac{1}{24} + \lim_{x\to 0 } \frac{ 2 \sin\left( \frac{x^3}{2(1-x^2)} \right) \sin\left( \frac{x(2-x^2)}{1-x^2} \right) }{ x^4 }$

$= - \frac{1}{24} + 2\lim_{x\to 0 } \frac{ \sin\left( \frac{x^3}{2(1-x^2)} \right) }{x^3} \lim_{x\to 0 } \frac{\sin\left( \frac{x(2-x^2)}{2(1-x^2)} \right) }{ x }$

$= - \frac{1}{24} + 2 ~ (\frac{1}{2}) ~( 1 ) = 1 - \frac{1}{24} = \frac{23}{24}$

**Ted** · June 24th 2010, 04:11 AM

Utherr :
Thanks. I will try it.

simplependulum:
Thanks. But sorry, I do not know what does the symbol O mean.

**Utherr** · June 24th 2010, 04:14 AM

double post :-s

**simplependulum** · June 24th 2010, 04:15 AM

Originally Posted by Ted

Utherr :
Thanks. I will try it.

simplependulum:
Thanks. But sorry, I do not know what does the symbol O mean.

it doesn't matter , it means the order of x is 6 so O(x^6) tends to zero when x does too (6 > 4 ) .

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