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Let $f:[0,\infty)\to\mathbb{R} \ $ be a continuous function.

Suppose $\int_{0}^{\infty}f(x)dx\ $ converges,

Compute: $$\lim\limits_{n\to\infty}\int_{0}^{1}f(nx)dx.$$

My attempt:

I took $t=nx$, then $dt=n\ dx$, so we get:

$$\lim\limits_{n\to\infty}\int_{0}^{1}f(nx)dx=\lim\limits_{n\to\infty}\frac{1}{n}\int_{0}^{n}f(t)dt=0\cdot\int_{0}^{\infty}f(x)dx=0.$$

Is it true? Am I missing something?

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    Looks ok to me. – 2017-02-04
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    As long as you know what you're doing at the last step, it looks good. – 2017-02-04
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    What do you mean? @OpenBall – 2017-02-04
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    @TrueTopologist **if** both limits exist (in this case they do) it implies limit of product is product of limits. – 2017-02-04
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    The only strange part is that $f$ is continuous. I do not see why this should be. – 2017-02-04
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    all you need is for $f$ to be $L^1$. – 2017-02-04
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    @AdamHughes yes I know this as a fact from a prior course. – 2017-02-04
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    TrueTopologist yes presumably so, I was just answering the question you asked @OpenBall. It wasn't clear from what you had written, hence his question. – 2017-02-04

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