5.27. The function $f(x)$ is not defined at $x=0$. Determine the value of $f(0)$ so that $f(x)$ is continuous at $x=0$, if

$$
f(x)=\frac{\sqrt[3]{1+x}-1}{\sqrt{4+x}-2}
$$
Solution. Let's find the limit of $f(x)$ at the point $x=0$. We have

$$
\begin{gathered}
\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-1}{\sqrt{4+x}-2}= \\
=\lim _{x \rightarrow 0} \frac{(\sqrt[3]{1+x}-1)(\sqrt{4+x}+2)\left(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}+1\right)}{(\sqrt{4+x}-2)(\sqrt{4+x}+2)\left(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}+1\right)}= \\
=\lim _{x \rightarrow 0} \frac{(1+x-1)(\sqrt{4+x}+2)}{(4+x-4)\left(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}+1\right)}= \\
=\lim _{x \rightarrow 0} \frac{\sqrt{4+x}+2}{\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}+1}=\frac{4}{3}
\end{gathered}
$$

The function will be continuous at the point $x=0$ if we set $f(0)=4 / 3$.

## 5.9 .  PROBLEMS

## CALCULATION OF THE LIMIT

 OF A NUMERICAL SEQUENCE5.28. $\lim _{x \rightarrow \infty} \frac{2 n^{2}+1}{3 n^{2}}$