Computing Limits of Multivariable Functions I have seen this problem on an exam I was looking at:
Let $f(x,y) = \frac{y}{1+xy} - \frac{1 - y\sin(\frac{\pi x}{y})}{\arctan(x)}$
Compute the limit:
$g(x) = \lim_{y\to\infty} f(x,y)$
I am confused on how to solve this limit however. Wouldn't the value of the limit also depend on $x$ ? (this problem lets $x$ be independent?)
 A: Let $f(x,y) = \frac{y}{1+xy} - \frac{1-ysin(\frac{\pi x}{y})}{\arctan(x)}$, let's modify our function
$$\frac{y}{1+xy} - \frac{1-ysin(\frac{\pi x}{y})}{\arctan(x)} = \frac{1}{x}\frac{1}{\frac{1}{xy}+1} - \frac{1-ysin(\frac{\pi x}{y})}{\arctan(x)},$$
now let $z = \frac{1}{y}$, hence
$$\lim_{y\rightarrow \infty} f(x,y) = \lim_{z\rightarrow 0} f(x,\frac{1}{z})=\lim_{z\rightarrow 0}\frac{1}{x}\frac{1}{\frac{z}{x}+1} - \frac{1-\frac{1}{z}sin(z\pi x)}{\arctan(x)},$$
clearly $\frac{1}{x}\frac{1}{\frac{z}{x}+1} \rightarrow \frac{1}{x},$
let's have a closer look at $\frac{1-\frac{1}{z}sin(z\pi x)}{\arctan(x)}$.
Using Taylor series we can write 
$$\frac{1}{z}sin(z\pi x) = \frac{1}{z}\Big(z\pi x + \mathcal{O}(z^3)\Big)=\pi x + \mathcal{O}(z^2),$$
Can you go from here?
A: HINT:
Recall that the sine function satisfies the inequalities
$$\left|z-\frac16 z^3\right| \le |\sin(z)|\le |z|$$
Now let $z=\pi x/y$.
A: Since you are having problems with the second term,
note that you can write it as
$$\frac{1-\frac{\sin\pi xp}{p}}{tan^{-1}x}$$
(here $\frac{1}{p}=y$ and as y goes to $\infty$, p goes to $0$) leaving the $tan^{-1}x$ alone,
our job is to find
$$\lim_{p\to 0} \frac{\sin\pi xp}{p}$$=
$$\lim_{p\to 0} \frac{\sin\pi xp}{\pi xp}\cdot \pi x$$
=$$\pi x$$
