30. Find the limits of the function $y=\frac{5}{2-x}$: 1) as $x \rightarrow 2-0$ and 2) as $x \rightarrow 2+0$. Explain the solution using tables.
Solution. 1) If $x$ tends to 2 from the left, remaining less than 2, then $2-x$ will be a positive infinitesimal,
and $\frac{5}{2-x}$ will be a positive infinitely large, i.e., if $x \rightarrow 2-0$, then $(2-x) \rightarrow+0$, and $\frac{5}{2-x} \rightarrow+\infty$, so $\lim _{x \rightarrow 2-0} \frac{5}{2-x}+\infty$.

The behavior of the variables $x, 2-x+\frac{5}{2-x}$ is explained by the following table:

| $x$ | 1 | 1.9 | 1.99 | 1.999 | 1.9999 | 1.99999 | 1.999999 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $2-x$ | 1 | 0.1 | 0.01 | 0.001 | 0.0001 | 0.00001 | 0.000001 | $\ldots$ |
| $\frac{5}{2-x}$ | 5 | 50 | 500 | 5000 | 50000 | 500000 | 5000000 | $\cdots$ |

2) If $x \rightarrow 2+0$, then $(2-x) \rightarrow 0$, and $\frac{5}{2-x} \rightarrow -\infty$, i.e., $\lim _{x \rightarrow 2+0} \frac{5}{2-x}=-\infty$.

The table of corresponding values of the variables $x, 2-x$, and $\frac{5}{2-x}$ vividly shows their behavior:

| $x$ | 3 | 2.1 | 2.01 | 2.001 | 2.0001 | 2.00001 | 2.000001 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $2-x$ | -1 | -0.1 | -0.01 | -0.001 | -0.0001 | -0.00001 | -0.000001 | $\ldots$ |
| $\frac{5}{2-x}$ | -5 | -50 | -500 | -5000 | -50000 | -500000 | -5000000 | $\ldots$ |
|  |  |  |  |  |  |  |  |  |

The graph of the function $y=\frac{5}{2-x}$ is shown in Fig. 16.