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Mathematics 16 Online
OpenStudy (anonymous):

limx→π/4cos2x÷sinx−cosx

OpenStudy (anonymous):

Is it as x approaches pi?

OpenStudy (anonymous):

\[\lim_{x \rightarrow \pi} (4\cos 2x)\div(\sin x -\cos x)\]

OpenStudy (anonymous):

probably he meant: \[\lim_{x \rightarrow \pi/4}\] Please confirm

OpenStudy (anonymous):

True. Good call.

OpenStudy (anonymous):

i am going to guess it is \[\lim_{x \to \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x)-\cos(x)}\]so you would get \[\frac{0}{0}\] by substitution.

OpenStudy (anonymous):

but it looks like it is a one step l'hopital problem if that is the question

OpenStudy (anonymous):

\[\lim_{x\to \frac{\pi}{4}}\frac{2\sin(2x)}{\cos(x)+\sin(x)}\] and done

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