Question

convergence test
(|cos x|)/(x^2-|sin x|) dx limits 1 to infinity

Answer 1

limits 1 to infinity
converge or diverge and proof
(|cos x|)/(x^2 - |sin x|)

Answer 2

The absolute value of cosine is always less than or equal to one. So you may take,\[\frac{|\cos x|}{x^2 - |\sin x|} \le \frac{1}{x^2- |\sin x|}=\frac{1/x^2}{1-\frac{|\sin x|}{x^2}}\]

Answer 3

\[\lim_{x \rightarrow \infty}\frac{|\cos x|}{x^2 - |\sin x|} \le \lim_{x \rightarrow \infty}\frac{1/x^2}{1-\frac{|\sin x|}{x^2}}\]\[=\frac{\lim_{x \rightarrow \infty}1/x^2}{\lim_{x \rightarrow \infty}1-\lim_{x \rightarrow \infty}\frac{|\sin x|}{x^2}}=\frac{0}{1-0}=0\]

Answer 4

Damn, just spotted the 'dx'. I thought you wanted to see what the limit of this thing was.

Are you using the integral test?