My Math Forum simplification of a limit

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 January 11th, 2013, 02:06 PM #1 Newbie   Joined: Jan 2013 Posts: 13 Thanks: 0 simplification of a limit Hi, I've a problem with the semplification of a limit.. My limit is: $\lim_{x \to 0} \left( \frac{-sinx+cosx\cdot x}{2x^2\cdot sinx} \right)= -\frac{1}{6}$ I know the result because I used wolframalpha :P then, I've think to divide this limit in this way: $\lim_{x \to 0} -\frac{1}{2x^2} + \lim_{x \to 0} \frac{x}{sinx} \cdot \lim_{x \to 0} \frac{cosx}{2x^2}$ I've divided the fraction and I've used the properties of limits.. There is: $\lim_{x \to 0} \frac{x}{sinx}= 1$ So I've think to omit it and rewrite: $\lim_{x \to 0} -\frac{1}{2x^2} + \lim_{x \to 0} \frac{cosx}{2x^2}= \lim_{x \to 0} \left( -\frac{1}{2x^2} + \frac{cosx}{2x^2} \right)$ but: $\lim_{x \to 0} \left( -\frac{1}{2x^2} + \frac{cosx}{2x^2} \right)= -\frac{1}{4}$ I think that I can't resolve only x/sinx.. why? Can someone explain me the invalid passage please? Thanks!
January 11th, 2013, 03:45 PM   #2
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Re: simplification of a limit

Hello, elviro!

Quote:
 $\text{Show that: }\;\lim_{x \to 0} \left( \frac{x\,\!\cos x\,-\,\sin x}{2x^2\,\!\sin x} \right) \:=\: -\frac{1}{6}$

If you are allowed L'Hopital's rule, apply it twice.

 January 12th, 2013, 02:06 AM #3 Newbie   Joined: Jan 2013 Posts: 13 Thanks: 0 Re: simplification of a limit yes, but I don't understand why my method isn't right
January 12th, 2013, 12:29 PM   #4
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Re: simplification of a limit

Quote:
 Originally Posted by elviro yes, but I don't understand why my method isn't right
The x/sinx term can't simply be repalced by 1. You need to include the next term in the expansion (x^2/6) when taking the limit.

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