Example 6. Find the one-sided limits of the function $f(x)=\frac{6}{x-3}$ as $x \rightarrow 3$ from the left and from the right.
Solution. The problem reduces to finding two limits

$$
\lim _{x \rightarrow 3-0} \frac{6}{x-3} \text { and } \lim _{x \rightarrow 3+0} \frac{6}{x-3}
$$

If $x \rightarrow 3-0$, i.e., $x$ approaches 3 while remaining less than 3, then the quantity $x-3$ is an infinitesimally small negative value. Its reciprocal will be infinitely large and also negative, and the same property applies to the quantity $y=\frac{6}{x-3}$. Therefore,

$$
\lim _{x \rightarrow 3-0} \frac{6}{x-3}=-\infty
$$

The changes in the variables $x, x-3$, and $\frac{6}{x-3}$ as $x \rightarrow 3-0$ can be illustrated by the following table:

| $x$ | 2 | 2.9 | 2.99 | 2.999 | 2.9999 | 2.99999 | 2.999999 |
| :---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| $x-3$ | -1 | -0.1 | -0.01 | -0.001 | -0.0001 | -0.00001 | -0.000001 |
| $\frac{6}{x-3}$ | -6 | -60 | -600 | -6000 | -60000 | -600000 | -6000000 |

If $x \rightarrow 3+0$, i.e., $x$ approaches 3 while remaining greater than 3, then the quantity $x-3$ is a positive infinitesimally small value. Its reciprocal $\frac{1}{x-3}$ will be infinitely large and positive. The same property applies to the quantity $\frac{6}{x-3}$, therefore,

$$
\lim _{x \rightarrow 3+0} \frac{6}{x-3}=+\infty
$$

The graph of the function $y=\frac{6}{x-3}$ is shown in Figure 1.22.

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-036.jpg?height=521&width=677&top_left_y=55&top_left_x=70)

Figure 1.22