115. Find $\lim _{x \rightarrow 3} \frac{x^{2}-9}{3-x}$.
Solution. A direct transition to the limit is impossible here, since the limit of the denominator is zero: $\lim _{x \rightarrow 3}(3-x)=3-3=0$. The limit of the dividend is also zero: $\lim _{x \rightarrow 3}\left(x^{2}-9\right)=9-9=0$. Thus, we have an indeterminate form of $0 / 0$. However, this does not mean that the given function does not have a limit; to find it, we need to preliminarily transform the function by dividing the numerator and the denominator by the expression $x-3$:

$$
\frac{x^{2}-9}{3-x}=\frac{(x-3)(x+3)}{3-x}=-\frac{(x-3)(x+3)}{x-3}=-(x+3)
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-179.jpg?height=334&width=379&top_left_y=187&top_left_x=118)

Fig. 95

$$
\begin{aligned}
& \text { For the expression }-(x+3) \text {, the limit as } \\
& x \rightarrow 3 \text { is easily found: } \lim _{x \rightarrow 3}(-(x+3))= \\
&=-\lim _{x \rightarrow 3}(x+3)=-6 .
\end{aligned}
$$

Remark. By canceling the fraction by $x-3$, we assume that $x \rightarrow 3$, but $x \neq 3$.