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I'm working through some trig questions and don't understand this one:
$\displaystyle \lim_{ x \to 0} \frac{1 - \cos 2x}{3x^{2}}$
My first thought was since $\displaystyle \lim_{x \to 0} \cos x = 1$, the answer would be 0, but alas, it turned out to be $\displaystyle \frac{2}{3}$
Could anyone shed some light on this?
Note that $\displaystyle \cos (2x)=1-2\sin^2(x)$ thusQuote:
Originally Posted by Lepzed
$\displaystyle \frac{1 - \cos (2x)}{3x^{2}}=\frac{2\sin^2(x)}{3x^{2}}=\frac{2}{3 }\left(\frac{\sin(x)}{x}\right)^2$
Ah, so I should've rewritten the cos 2x expression, before applying the limit.
Thanks! Really helpful!
I worked through another exercise and I'm wondering if my approach is correct. It might be a bit verbose, but I'd like to be clear on my train of thought:
$\displaystyle \lim_{x \to 0} \frac{x^2}{\sin x \tan 3x} = \lim_{x \to 0} \frac{x^2}{\sin x \frac{\sin 3x}{\cos 3x}} = \lim_{x \to 0} \frac{x^2}{\frac{ \sin x \cos x \sin 3x}{\cos 3x}} = \lim_{x \to 0} \frac{x^2 \cos 3x}{\sin x \cos x \sin 3x} = \lim_{x \to 0} \frac{x \cos 3x}{ \sin 3x \cos x } = \lim_{x \to 0} \frac{x}{ \sin 3x} = \frac{1}{3}$
Am I doing anything illegal here? I can't really identify with the meaning of this, so it's a mechanical process for me, I can't visualize this at all.
It would easier to follow writing it as:Quote:
Originally Posted by Lepzed
$\displaystyle \left( {\frac{x}{{\sin (x)}}} \right)\left( {\frac{x}{{\sin (3x)}}} \right)\left( {\frac{{\cos (3x)}}{{\cos (x)}}} \right)$
