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Mathematics 13 Online
OpenStudy (watchmath):

Without L'hospital, compute \[\lim_{x\to 0}\frac{\cos(\frac{\pi}{2}\cos x)}{\sin(\sin x)}\]

OpenStudy (zarkon):

use \[\cos(\frac{\pi}{2}\cos(x))=\sin\left(\frac{\pi}{2}\cos(x)+\frac{\pi}{2}\right)\]

OpenStudy (watchmath):

you mean sin(pi/2 - (pi/2)cosx)?

OpenStudy (zarkon):

you can use that too.... then use the fact that \[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

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