using L'Hospital's rule, solve lim x-0 cot2x*sin6x and lim x-infinity x-lnx. I know I need common denominators....can't figure out how to get them.

Question

using L'Hospital's rule, solve lim x->0 cot2x*sin6x and lim x->infinity x-lnx. I know I need common denominators....can't figure out how to get them.

anonymous · Answer

$$=\lim [\cos(2x)*\sin6x]/\sin(2x)=
\[=\lim(-2\sin2x*\sin6x+6\cos2x*\cos6x)/2\cos2x=6/2=3$$

myininaya · Answer

$$\lim_{x \rightarrow 0}\frac{\cos(2x)}{\sin(2x)}*\sin(6x)=\lim_{x \rightarrow0}\frac{\cos(2x)*\sin(6x)}{\sin(2x)}$$
we have 0/0 so we can apply lhospital
$$=\lim_{x \rightarrow 0}\frac{-2\sin(2x)*\sin(6x)+6\cos(6x)*\cos(2x)}{2\cos(2x)}$$

myininaya · Answer

$$=\frac{-2(0)*0+6*1*1}{2*1}=\frac{6}{2}=3$$

anonymous · Answer

dude u rock. i thought i couldn't do that for some reason, so i'm glad I double checked. good to know i can trust my own instincts about this stuff tho.