Question

how do you find the limit of: inv sin (x/(1-2x)) as x approaches positive infinite?

Answer 1

Is this the problem?\[\lim_{x \rightarrow \infty} \sin^{-1}\left(\frac{x}{1-2x}\right)\]

Answer 2

If so, you would need to eliminate the x's such that the inverse sin's argument is not positive or negative infinity or undefined:\[\lim_{x \rightarrow \infty} \sin^{-1}\left(\frac{x \times\frac{1}{x}}{(1-2x) \times\frac{1}{x}}\right)\]\[\lim_{x \rightarrow \infty} \sin^{-1}\left(\frac{1}{\frac{1}{x}-2}\right) \]Now you can evaluate it as x goes to infinity:\[\sin^{-1}\left(\frac{1}{\frac{1}{\infty}-2}\right) = \sin^{-1}\left(\frac{1}{0-2}\right)\] \[=\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \]