Limit Problem (L'Hopital's Rule)

**Warrenx** · December 4th 2010, 02:34 PM

Edit: I Think the equation came out wrong in latex. Here is the updated one.

34. $lim_{x\to\infty}\frac{sin^2x - 2x}{x^2sin^2x}$

I have done it out, and for the numerator I get $2cos^2x - 2sin^2x -2$ and I just know it wants to be a pythagorean and cancel out leaving me with 0. I think I am messing up somewhere because I took it to the second derivation and it just gets funny after that.

Any help would be appreciated.

**FernandoRevilla** · December 4th 2010, 02:56 PM

The limit does not exist.

(i) $\displaystyle\lim_{x \to{+}\infty}{(1/x^2)}=0$

However, choosing:

$x_n=\dfrac{\pi}{2}+2n\pi,\;y_n=\dfrac{\pi}{4}+2n\p i\;\;(n\in\mathbb{N})$

we have:

(ii) $\displaystyle\lim_{n \to{+}\infty}{\csc^2 x_n}=1,\;\displaystyle\lim_{n \to{+}\infty}{\csc^2 y_n}=2$

Regards.

Fernando Revilla

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