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Find the limit as x approaches 0 for (t^2)/(〖1-cos〗^2 t)
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I think it should be \[\lim_{t \rightarrow 0} \frac{t^2}{(1 - \cos(?))^2 t}\] but if its as above you can remove a common factor of t from the numerator and denominator.
its not that that one haha its \[\frac{ t ^{2} }{ 1-\cos ^{2}t }\]
\[\huge \frac{t^2}{\sin^2t}=\frac{1}{(\sin^2t )\times \frac{1}{t^2}}=\frac{1}{\frac{\sin^2t}{t^2}}\] \[\lim_{t \rightarrow 0} \frac{\sin t}{t}=0\]
the answer is suppose to be 1...
sry \[\frac{\sin t}{t}=1\] \[\frac{1}{1^2}=1\] yes
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oh i see it, thanks!
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