Evaluate $\lim_{x \to \infty}\left(\frac{2}{\pi}(x+1)\cos^{-1}\left(\frac{1}{x}\right)-x\right)$ My try: Since , $x \to \infty$ , $\dfrac{1}{x} \to 0$ , and therefore , $\cos^{-1}\left(\dfrac{1}{x}\right) \to \dfrac{\pi}{2}$.
This however , gives me the answer as $1$ , which is not correct. Can anyone tell me where I messed up ? 
Someone told me I cannot substitute the values of $x$ directly when I have an $\infty-\infty$ form . Why is it true ? And how should I proceed next then ?
 A: Set $x=\dfrac1h$ to find
$$\dfrac2\pi\lim_{h\to0^+}\left(\dfrac{(1+h)\arccos(h)-\dfrac\pi2}h\right) =\dfrac2\pi\lim_{h\to0^+}\left(\arccos(h)-\dfrac{\arcsin h}h\right)$$  as Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$
A: As lab bhattacharjee answered, start defining $x=\frac 1y$ to make 
$$A=\frac{2 }{\pi }(x+1) \cos ^{-1}\left(\frac{1}{x}\right)-x=\frac{2 (y+1) \cos ^{-1}(y)-\pi }{\pi  y}$$ Now, using Taylor series 
$$\cos ^{-1}(y)=\frac{\pi }{2}-y+O\left(y^3\right)$$ which make
$$A=\frac{2(y+1)\left( \frac{\pi }{2}-y+O\left(y^3\right)\right)-\pi } {\pi y }=\frac{(\pi -2) y-2 y^2+O\left(y^3\right) }{\pi y  }=\frac{\pi -2}{\pi }-\frac{2 y}{\pi }+O\left(y^2\right)$$ which shows the limit when $y\to 0$ and how it is approached.
Back to $x$
$$A=\frac{2 }{\pi }(x+1) \cos ^{-1}\left(\frac{1}{x}\right)-x=\frac{\pi -2}{\pi }-\frac{2 }{\pi x}+O\left(\frac{1}{x^2}\right)$$ Just for your curioisity, graph both functions on the same plot and notice hoaw close they are even for $x=10$. The exact value would be $\approx 0.2985$ while the approximation would give $\approx 0.2997$.
A: $$\lim_{x\rightarrow \infty}\bigg[\frac{2}{\pi}\bigg((x+1)\frac{\pi}{2}-(x+1)\arcsin\bigg(\frac{1}{x}\bigg)\bigg)-x\bigg]$$
So $$1-\lim_{x\rightarrow 0}\frac{2}{\pi}\frac{\arcsin(x)}{x}$$
So  limit is $$1-\frac{2}{\pi}$$
