I can't seem to find out where to start or finish for this limit: Lim 1-cos 3x / cos(squared) 3x - 1 x->0
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Just factorise the bottom.
sorry, that would make it a lot easier, but the bottom 3x was suppose to be a 5x. Thats what makes it so complicated. NEED HELP Lim 1-cos 3x / cos(squared) 5x - 1 x->0
$\displaystyle \frac{{1 - \cos 3x}} {{\cos ^2 5x - 1}} = - \frac{9} {{25}} \cdot \frac{{1 - \cos 3x}} {{9x^2 }}\left( {\frac{{5x}} {{\sin 5x}}} \right)^2 .$ The answer should be obvious from there.
that doesn't make it obvious. could you explain a little farther. Thanks
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