lim x->0 (cot2x)/(cscx) how do you solve this?
I would first write in terms of sin and cos I most always do that with limits involving trig
\[\text{ I would also recall these limits: } \lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 \text{ and } \lim_{x \rightarrow 0} \frac{\cos(x)-1}{x}=0\] There some other things to recall to but impossible to write a whole book on limits here
i'd like a step by step of how to do it because i get confused once i turn cot2x into cos2x/sin2x
what is csc(x)?
\[\lim_{x \rightarrow 0}\frac{\cot(2x)}{\csc(x)}=\lim_{x \rightarrow 0}\frac{\frac{\cos(2x)}{\sin(2x)}}{\frac{1}{\sin(x)}}\] I hate compound fractions so I attempt to get rid of them when I see them. Do you know how to do that? (And I went ahead and answered the question I directed to you) if you know what cot is in terms of cos and sin then you should know what csc is which is 1/sin
yes i've gotten this far but i don't know what to do after this.
so you know how to write as a simple fraction then? is so how?
recall \[\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{\frac{a}{b}}{\frac{c}{d}} \cdot \frac{\frac{d}{c}}{\frac{d}{c}} \text{ (Note (d/c)/(d/c)=1) } \\ =\frac{\frac{a}{b} \cdot \frac{d}{c}}{1} =\frac{ad}{bc}\]
notice we just went from a compound fraction to a simple fraction
if you don't know how to use latex at least draw for me what we have in our problem you know writing that compound fraction as a simple fraction
\[(\cos2x(sinx))/\sin2x\]
is that right?
beautiful
i like to apply the first sin limit I mentioned whenever I can \[\lim_{u(x) \rightarrow 0}\frac{\sin(u(x))}{u(x)}=1\] let u be a function of x (but notice I'm rewriting it here a little mostly because of the 2x part) now the reciprocal of that limit is also 1 we could verify this with the squeeze theorem (but this is not really called for here)
Let me write our fraction out better so I think you can see better how to apply this limit
|dw:1404955629711:dw| I'm going to put little squares here (rectangles whatever) I want you to tell me which goes in each of those rectangles so we could either have something like sin(u)/u or u/sin(u)
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