Saturday, 3 June 2017

calculus - How to calculate limxtoinftyfracxxsinx?


I tried to solve
limxxxsinx.




After dividing by x I got that it equals to:
limx11sinxx.

Now, using L'hopital (0/0) I get that
limxsinxx=limxcosx

and the lim at infinity for cosx is not defined. So basically I get that the overall limit of
limxxxsinx
is 1 or not defined?

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