142. Find $\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})$.
Solution. Here, it is required to find the limit of the difference of two quantities tending to infinity (indeterminate form of $\infty-\infty$). By multiplying and dividing the given expression by its conjugate, we get

$$
\sqrt{x+1}-\sqrt{x}=\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}=\frac{1}{\sqrt{x+1}+\sqrt{x}}
$$

Therefore,

$$
\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})=\lim _{x \rightarrow \infty} \frac{1}{\sqrt{x+1}+\sqrt{x}}=0
$$

Let's consider examples where remarkable limits are used.