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{\bf Question}

Consider the sequence $\{f_n\}$ of functions

$\ds f_n(x)=\left\{\begin{array}{cl}n^2x&amp;0\leq x\leq\frac{1}{n}\\
\\ 2n-n^2x&amp; \frac{1}{n}&lt;x\leq\frac{2}{n}\\ \\0&amp; {\rm
otherwise}\end{array}\right.$

${}$

Evaluate $\int\lim f_n$ and $\lim\int f_n$ and comment.



\vspace{0.25in}

{\bf Answer}

$\ds\lim_{n\to\infty}f_n(x)=0$ therefore $\ds \int\lim f_n=0$

\begin{eqnarray*}\int f_n &amp;=&amp;
\int_0^{\frac{1}{n}}n^2x+\int_{\frac{1}{n}}^{\frac{2}{n}}2n-n^2x\\
&amp;=&amp; n^2\left[\frac{x^2}{2}\right]_0^{\frac{1}{n}}+
\left[2nx-n^2\frac{x^2}{2}\right]_{\frac{1}{n}}^{\frac{2}{n}}\\
&amp;=&amp;
n^2\frac{1}{2}\frac{1}{n^2}+2n\frac{1}{n}-\frac{n^2}{2}\frac{3}{n^2}\\
&amp;=&amp; \frac{1}{2}+2-\frac{3}{2}=1\end{eqnarray*}

Therefore $\lim f_n=1$


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