Example 4. Find $\lim _{y \rightarrow 0} \frac{e^{y}+\sin y-1}{\ln (1+y)}$.
Since $\lim _{y \rightarrow 0} \frac{e^{y}-1}{y}=1, e^{y}-1 \sim y$
$$
\lim _{\substack{(y \rightarrow 0 \\(y \rightarrow 0 \text { when) }}} \frac{e^{y}-1+\sin y}{2 y}=1, e^{y}-1+\sin y \sim 2 y
$$

Therefore, the original expression $=\lim _{y \rightarrow 0} \frac{2 y}{y}=2$.
It is worth noting that when using equivalent infinitesimal substitution, only local equivalent infinitesimal substitution is performed, without overall verification, which may lead to incorrect results.

For example, when finding $\lim _{x \rightarrow 0} \frac{\operatorname{tg} x-\sin x}{x^{3}}$, only noticing $\sin x \sim x$ and $\operatorname{tg} x \sim x (x \rightarrow 0$ when $)$, and making the following substitution:
Original expression $=\lim _{x \rightarrow 0} \frac{x-x}{x}=0$, this result is incorrect.
In fact, the original expression $=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{\left(\frac{1}{\cos x}-1\right)}{x^{2}}$
$$
=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} \cdot \frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}=\frac{1}{2} \text {. }
$$

In finding limits, using equivalent infinitesimals for substitution is often effective.

Let $\lim \alpha(x)=\infty, \lim \beta(x)=\infty$ $\left(x \rightarrow x_{0}\right.$ or $x \rightarrow \infty$ when $)$, if $\lim \frac{\beta(x)}{\alpha(x)}=1$, then $\beta(x)$ and $\alpha(x)$ are said to be equivalent infinitesimals, denoted as $a(x) \sim \beta(x) \quad\left(x \rightarrow x_{0}\right.$ or $x \rightarrow \infty$ when $)$.

Similarly, $\lim \frac{\beta(x)}{\alpha(x)}=\lim \frac{\beta_{1}(x)}{\alpha_{1}(x)}$ $\left(\alpha(x) \sim a_{1}(x), \beta(x) \sim \beta_{1}(x)\right.$, when $x \rightarrow 0$ or $x \rightarrow \infty$).
For example, $\lim _{x \rightarrow+\infty} \frac{e^{x}-e^{-2 x}}{3^{x}+\frac{2}{-x}}$.
Since $e^{x} \sim e^{x}-e^{-2 x}, \quad 3^{x} \sim 3^{x}+2^{-x}$ $(x \rightarrow+\infty$ when $)$, the original expression $=\lim _{x \rightarrow+\infty} \frac{e^{x}}{3^{x}}$ $=\lim _{x \rightarrow+\infty}\left(\frac{e}{3}\right)^{x}=0$.
(Author's unit: Department of Mathematics, Tianjin Normal University)