Saturday, 3 June 2017

calculus - How to calculate limxtoinftyfracxxsinx?


I tried to solve
lim




After dividing by x I got that it equals to:
\lim_{x\to\infty}\frac{1}{1-\frac{\sin x}{x}}. Now, using L'hopital (0/0) I get that
\lim_{x\to\infty}\frac{\sin x}{x} = \lim_{x\to\infty}\cos x
and the lim at infinity for \cos x is not defined. So basically I get that the overall limit of
\lim_{x\to\infty}\frac{x}{x-\sin x} is 1 or not defined?

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