Trig Limits (1 Viewer)

Sleight_of_Hand · Jul 25, 2009

lim theta approaches 0

theta^2 / 1 - sin^2 theta

thanks for the help

Drongoski · Jul 25, 2009

If I'm not mistaken:

$\100dpi \lim_{\theta \to 0}\ \frac{\theta^{2} }{1-sin^{2}\theta }\ \ =\ \ \frac{\lim_{\theta \to 0}\theta ^{2}}{\lim_{\theta \to 0}(1-sin^{2}\theta )}\ \ =\ \ \frac{0^{2}}{1\ -\ 0^{2}}\ \ =\ \ 0$

Trig Limits (1 Viewer)

Sleight_of_Hand

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Drongoski

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