$f(x)$ 是定义在 $[0,+\infty)$ 上的函数, 具体为
\[f(x)=\lim\limits_{n\rightarrow\infty}\sqrt[n]{1+x^n+(\frac{x^2}{2})^n},\]
证明 $f(x)\in C([0,+\infty))$.
1
[Idea] 先求特殊点处的值, 然后分段求出 $f(x)$, 事实上所得的 $f(x)$ 是一个分段函数.
当 $x=0$ 时,
\[f(0)=\lim_{n\rightarrow\infty}\sqrt[n]{1+0^n+(\frac{0^2}{2})^n}=\lim_{n\rightarrow\infty}\sqrt[n]{1}=1.\]
当 $x=1$ 时,
\[f(1)=\lim_{n\rightarrow\infty}\sqrt[n]{1+1^n+(\frac{1^2}{2})^n}=\lim_{n\rightarrow\infty}\sqrt[n]{2+\frac{1}{2^n}},\]
注意到
\[\sqrt[n]{2} < \sqrt[n]{2+\frac{1}{2^n}} < \sqrt[n]{3},\]
而 $\lim\limits_{n\rightarrow\infty}\sqrt[n]{a}=1$, 对任意 $a > 0$ 成立, 故 $\lim\limits_{n\rightarrow\infty}\sqrt[n]{2+\frac{1}{2^n}}=1$.
当 $x=\sqrt{2}$ 时,
\[f(\sqrt{2})=\lim_{n\rightarrow\infty}\sqrt[n]{1+(\sqrt{2})^n+(\frac{(\sqrt{2})^2}{2})^n}=\lim_{n\rightarrow\infty}\sqrt[n]{2+(\sqrt{2})^n},\]
注意到
\[
\sqrt{2}=\sqrt[n]{(\sqrt{2})^n} < \sqrt[n]{2+(\sqrt{2})^n} < \sqrt[n]{2\cdot(\sqrt{2})^n}=\sqrt[n]{2}\cdot\sqrt{2},
\]
故 $\lim\limits_{n\rightarrow\infty}\sqrt[n]{2+(\sqrt{2})^n}=\sqrt{2}$.
当 $x\in(0,1)$ 时,
\[\sqrt[n]{1} < \sqrt[n]{1+x^n+(\frac{x^2}{2})^n} < \sqrt[n]{3},\]
故 $\lim\limits_{n\rightarrow\infty}\sqrt[n]{1+x^n+(\frac{x^2}{2})^n}=1$.
当 $x\in(1,\sqrt{2})$ 时,
\[x=\sqrt[n]{x^n} < \sqrt[n]{1+x^n+(\frac{x^2}{2})^n} < \sqrt[n]{1+x^n+1}=\sqrt[n]{2+x^n} < \sqrt[n]{2\cdot x^n}=\sqrt[n]{2}\cdot x,\]
故 $\lim\limits_{n\rightarrow\infty}\sqrt[n]{1+x^n+(\frac{x^2}{2})^n}=x$.
当 $x > \sqrt{2}$ 时,
\[\sqrt[n]{(\frac{x^2}{2})^n} < \sqrt[n]{1+x^n+(\frac{x^2}{2})^n} < \sqrt[n]{2\cdot(\frac{x^2}{2})^n}=\sqrt[n]{2}\cdot\frac{x^2}{2},\]
故 $\lim\limits_{n\rightarrow\infty}\sqrt[n]{1+x^n+(\frac{x^2}{2})^n}=\frac{x^2}{2}$.
综上,
\[
f(x)=\begin{cases}
1, & 0\leqslant x\leqslant 1,\\
x, & 1 < x\leqslant\sqrt{2},\\
\frac{x^2}{2}, & x > \sqrt{2}.
\end{cases}
\]