Finding limit of $f(x)$ using appropriate method Consider $f(x) = \frac{1}{1+ \cos(x)}$ . I want to find $\lim _{x\to \infty} f(x)$ using appropriate method.
My try : I'm really confused about limit of $\sin(x)$ and $\cos(x)$. Therefore I can't solve this problem.
 A: $\lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow \infty}\frac{1}{1+cosx}=\lim_{x\rightarrow \infty}\frac{1}{2\times cos^2(\frac{x}{2})}=\frac{1}{2}\lim_{x\rightarrow \infty}\frac{1}{cos^2(\frac{x}{2})}$ 
which has liminf $=\frac{1}{2}$ and limsup $=\infty$ ( because $0\leq cos^2 \frac{x}{2}\leq 1$ ).  So doesn't converge but oscillates between $\frac{1}{2}$ and $\infty$.
A: Okay lets take $x=t+\pi$ then we have that
$$\lim_{x\to \infty}\frac{1}{1+\cos x}=\lim_{t\to \infty}\frac{1}{1+\cos (t+\pi)}=\lim_{t\to\infty}\frac{1}{1-\cos t}$$
So
$$\lim_{x\to\infty}\frac{1}{1+\cos x}=\lim_{x\to \infty}\frac{1}{1-\cos x}$$
This happens only if $\cos x\to 0$,now take $x=t-\pi/2$ and $x=t+\pi/2$ then we have
$$\lim_{t\to\infty}\frac{1}{1+\sin t}=\lim_{t\to\infty}\frac{1}{1-\sin t}$$Lets just change back the index to $x$ then we have that$$\lim_{x\to\infty}\frac{1}{1+\sin t}=\lim_{x\to\infty}\frac{1}{1-\sin t}$$
As before this implies that $\sin x\to 0$ but that's impossible when $\cos x\to 0$ hence contradiction the limit doesn't exist.
