lim x-> 0 x^2/(1-cos2x) lim x-> 0 x*(x/(1-cos2x)) lim x-> 0 (x/2)(2x/1-cos2x) lim x-> 0 x/2*0 =0/2*0 =0 just checking to see if i did the right thing
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Originally Posted by purplec16 lim x-> 0 x^2/(1-cos2x) lim x-> 0 x*(x/(1-cos2x)) lim x-> 0 (x/2)(2x/1-cos2x) lim x-> 0 x/2*0 =0/2*0 =0 That is not correct. This is ideal for using l'Hopiial's rule twice. If you don't have the use of l'HR then note that $\displaystyle \frac{{x^2 }}{{1 - \cos (2x)}} = \frac{1}{2}\left( {\frac{x}{{\sin (x)}}} \right)^2 .$
No we didnt learn L'Hospitals Rule
when you have a limit of the form 0/0 (as x--> a), you can't just call it "0" (such a limit is called "indeterminate"). hopefully, you have already covered what lim x-->0 sin(x)/x is.
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