<s>[INST] 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?)
 [/INST] 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?
</s>