using lim (radian ->0) (sin (radian))/(radian) = 1, find the limit of lim (t->0) (2t)/(tan t).
split tan into sin and cos and it's pretty straightforward
also you shouldn't write sin(radian), that doesn't make a lot of sense...
sine and cosine are functions of numbers, and as such they correspond to the right triangle trigonometry if you measure an angle in "radians" but if you write \(f(x)=x^2\) no one is going to ask you if \(x\) is in degrees, radians, Celsius, Fahrenheit or anything else. it is presumed to be a real number
\[\frac{2t}{\tan(t)}=\frac{2t\cos(t)}{\sin(t)}=2\times \frac{t}{\sin(t)}\times \cos(t)\] now take the limit piece by piece
sorry, that's how it was in the book, they called theta radians, but there's no place to add the symbol raidan when asking the original question, as you may have noticed. using \[\lim_{\theta \rightarrow 0} \frac{ \sin \theta }{ \theta }\] solve for \[\lim_{t \rightarrow 0} \frac{ 2t }{ \tan t }\] so I get \[\lim_{t \rightarrow 0} \frac{ 2t }{ \frac{ \sin t }{ \cos t } }\] Sorry, but it's not actually straight forward for me TuringTest, it's why I'm asking ^_^
satellite showed ou what I was trying to suggest to you\[\frac{2t}{\tan(t)}=\frac{2t\cos(t)}{\sin(t)}=2\times \frac{t}{\sin(t)}\times \cos(t)\]
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in case you have some doubt as to the limit of the middle term perhaps you may prefer it as\[\frac{2t}{\tan(t)}=\frac{2t\cos(t)}{\sin(t)}=2\times {\frac1{\sin t\over t}}\times \cos(t)\]
i would avoid writing compuund fractions if at all possible because they are confusing. instead of writing \[\frac{ 2t }{ \frac{ \sin t }{ \cos t } }\] it is best to go right to \[\frac{2t\cos(t)}{\sin(t)}\]
So you know how to evaluate the following limits? \[\lim_{t \rightarrow 0}\frac{t}{\sin(t)}=? ; \lim_{t \rightarrow 0}\cos(t)=?\] Are there any questions with what sat and turing said?
I believe we have lost our asker...
I believe the asker had to go to class. And I prefer writing out every single step, including compound fractions. It makes it clearer to me wtf went down in each step.
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