976. Find the following limits:
1) $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\tan x y}{y}$
2) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}$.
Solution. 1) $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\operatorname{tg} x y}{y}=\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} x \frac{\operatorname{tg} x y}{x y}=2 \cdot 1=2$,

since

$$
\lim _{\alpha \rightarrow 0} \frac{\operatorname{tg} \alpha}{\alpha}=1
$$

2) Let's use the definition of the limit of a function at a point. Take the sequences $x=\frac{1}{n} \rightarrow 0, y=\frac{1}{n} \rightarrow 0(n \rightarrow \infty)$,

then

$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}+\frac{1}{n}}{\frac{1}{n}}=2
$$

Now take $x=\frac{1}{n} \rightarrow 0, y=\frac{2}{n} \rightarrow 0(n \rightarrow \infty), \quad$ then

$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}+\frac{2}{n}}{\frac{1}{n}}=3
$$

Thus, for different sequences of points converging to the point $(0 ; 0)$, we get different limits. This means that the limit of the function at this point does not exist (see point 1).

Note that the given function is not defined at the point $(0 ; 0)$. Find the following limits: