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I need to show that this limit exists and then evaluate it. It is from a section on uniform convergence of sequences. I know that if $f_n \rightarrow f$ uniformly and each $f_n$ is integrable, then I can bring the limit inside of the integral. I'm not sure if this will be the right way to approach this, and if it is how to show that it is all of those things that would allow me to bring the limit in. Looking for some advice. Thanks

Felix Marin
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MFD55012
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2 Answers2

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Hint. If $0\le x\le 2$ then $$1\le e^{x^2/n}\le e^{4/n}.$$

Hanul Jeon
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  • The function is bounded for any n on the interval, but what do I do with that information? – MFD55012 Nov 04 '13 at 01:23
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    You can use this information to bound your sequence of integrals by a sequence of numbers which decreases and has limit 2. Your sequence is also bounded below by 2. Apply the squeeze theorem. – B. Mackey Nov 04 '13 at 01:31
  • The hint also tells you that the sequence is uniformly convergent as you asked for in the theorem you wanted to apply. – Keaton Nov 04 '13 at 02:05
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} \color{#0000ff}{\large\lim_{n \to \infty}\int_{0}^{2}\expo{x^{2}/n}\,{\rm d}x} &= \lim_{n \to \infty}\pars{\root{n}\int_{0}^{2/\!\root{n}}\expo{x^{2}}\,{\rm d}x} = \lim_{n \to \infty}{\ds{\int_{0}^{2n^{-1/2}}\expo{x^{2}}\,{\rm d}x} \over n^{-1/2}} \\[3mm]&= \lim_{n \to 0}{\ds{\int_{0}^{2n}\expo{x^{2}}\,{\rm d}x} \over n} = \lim_{n \to 0}{\expo{\pars{2n}^{2}}\pars{2} \over 1} = \color{#0000ff}{\large 2} \end{align}

Felix Marin
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