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
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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