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limit of t^2/1-cos^2t as t approaches zero. im lost I was going to use cost-1/t=0 but don't know how to get it
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\[\lim_{t \rightarrow 0}\frac{t^2}{1-\cos^2(t)}\]
so you know 1-cos^2(t)=sin^2(t) right?
\[\text{ use the following } \\ (g(x))^2=g(x) \cdot g(x) \\ \text{ and } \\ \lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1\]
actually the other way works to ... one sec
\[\lim_{x \rightarrow 0}\frac{x}{\sin(x)}=1 \]
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wow ive got to memorize trig identities. one question though how are you able to flip it? like how were you able to bring up sin sin^2?
\[\lim_{x \rightarrow 0}\frac{x}{\sin(x)} \cdot \frac{x}{\sin(x)}=1 \cdot 1 \]
this limit was proved by squeeze theorem
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