714. Find the limits:
1) $\lim _{\substack{x \rightarrow 3 \\ y \rightarrow 0}} \frac{\tan(x y)}{y}$
2) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x}{x+y}$.
Solution. Having made sure that the function is not defined at the limit point, we perform transformations, guided by the instructions in § 7, Chapter I:

1) $\lim _{\substack{x \rightarrow 3 \\ y \rightarrow 0}} \frac{\tan(x y)}{y}=\lim x \cdot \lim \frac{\tan(x y)}{x y}=3 \cdot 1=3$, since $\lim _{a \rightarrow 0} \frac{\tan \alpha}{a}=1$.
2) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x}{x+y}=\lim \frac{1}{1+\frac{y}{x}}$ - does not exist, because the ratio $\frac{y}{x}$

does not have a limit as the point $M(x, y)$ approaches the point $M_{0}(0 ; 0)$ arbitrarily. Thus, if $M \rightarrow M_{0}$ along different lines $y=k x$, then $\frac{y}{x}=k$, i.e., it depends on the slope of the line along which the point $M$ moves.