limx→0ln(1+sin3xcos2x)cot(ln3(1+x))tan4xsin(√x2+2−√2)ln(1+x2)
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerator). The given limit has a 00 form. I tried using taylor series but the it made the problem more complicated.
No comments:
Post a Comment