Limit of $\displaystyle \frac{1-\sin x}{1+\sin x}$ I have to find the limit of $\displaystyle \frac{1-\sin x}{1+\sin x}$ as $x\to\infty$. I was studying this function finding its real graph but to do this I need to know where the function goes as $x$ approaches to $\infty$. I just know that $\sin x$ as $x\to\infty$ is between -1 and 1 so is there a solution? How? Can I solve the limit using notable limits or de l'Hospital's rule?
Thanks in advance! (intuitively I would say that the solution is 0 probably)
 A: Let's suppose the limit exist or is plus or minus infinity. So we would have
$$\lim\limits_{x\to\infty}{1-\sin(x)\over 1+\sin(x)}=\alpha,$$
$\alpha \in \mathbb R\cup\{-\infty,+\infty\}$.
Now take $(x_n)=2\pi n$. Since $x_n \to \infty$ as $n \to \infty$, we would have 
$$\lim\limits_{n\to\infty}{1-\sin(x_n)\over 1+\sin(x_n)}=\alpha.$$
But $\sin(x_n)=\sin(2\pi n)=0$, so $\alpha = 1$.
Now take $(y_n) = 2\pi n + \pi/2$. Since $y_n \to \infty$ as $n \to \infty$, we would have too
$$ \lim\limits_{n\to\infty}{1-\sin(y_n)\over 1+\sin(y_n)}=\alpha.$$
But $\sin(y_n)=\sin(2\pi n+ \pi/2)=1$, so $\alpha = 0$. This is a contradiction, since, in $\mathbb R$, limits must be unique.
A: Since
$$\lim_{n\to\infty}\sin(2n\pi)=0\ne1= \lim_{n\to\infty}\sin\left(2n\pi+\frac \pi 2\right)$$
then
$$\lim_{x\to\infty}\sin x$$
doesn't exist.
Now assume that your given limit exist and equal to $\ell$ and find a contradiction.
A: The function is $2\pi$-periodic and nonconstant so the limit doesn't exist. 
Also, the function is not even well-defined for a sequence of $(x_n)$ with $x_n \to \infty$ as $n\to \infty$.  Most people would probably consider this an alternate sufficient reason for the limit not to exist.
