Example 6. Find $\lim _{x \rightarrow 3} \frac{\ln \left(x^{2}-5 x+7\right)}{x-3}$.
Solution. When $x \rightarrow 3$, we get an indeterminate form of $\frac{0}{0}$, since $x^{2}-5 x+7 \rightarrow 1$ and $\ln \left(x^{2}-5 x+7\right) \rightarrow 0$. The expression $x^{2}-5 x+7$ can be represented as:

$$
x^{2}-5 x+7=1+\left(x^{2}-5 x+6\right)=1+z
$$

where $\left(x^{2}-5 x+6\right)=z \rightarrow 0$ as $x \rightarrow 3$.

Since $\ln (1+z) \sim z$ as $z \rightarrow 0$ (see example 2), then

$$
\ln \left(x^{2}-5 x+7\right) \sim x^{2}-5 x+6
$$

Applying formula (1.74), we get

$$
\lim _{x \rightarrow 3} \frac{\ln \left(x^{2}-5 x+7\right)}{x-3}=\lim _{x \rightarrow 3} \frac{x^{2}-5 x+6}{x-3}=\lim _{x \rightarrow 3} \frac{(x-2)(x-3)}{x-3}=
$$

$$
=\lim _{x \rightarrow 3}(x-2)=1
$$