Example 3. Find $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}$.
Solution. Introduce a new variable $z$ by the formula

$$
a^{x}-1=z
$$

from which

$$
a^{x}=1+z
$$

Taking the logarithm of this equality to the base $e$, we obtain the formula

$$
x \ln a=\ln (1+z)
$$

from which

$$
x=\frac{\ln (1+z)}{\ln a}
$$

Obviously, as $x \rightarrow 0$ and $z \rightarrow 0$, therefore

$$
\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\lim _{z \rightarrow 0} \frac{z}{\frac{\ln (1+z)}{\ln a}}=\lim _{z \rightarrow 0} \frac{\ln a}{\frac{\ln (1+z)}{z}}
$$

Taking into account properties (1.44) and (1.46) and formula (1.65), we get

$$
\lim _{z \rightarrow 0} \frac{\ln a}{\frac{\ln (1+z)}{z}}=\ln \frac{z}{\lim _{z \rightarrow 0}\left[\frac{\ln (1+z)}{z}\right]}=\ln a
$$

Thus,

$$
\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\ln a
$$