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Can anyone help me with this? I want to know how to solve it.

Let $f:\mathbb R \longrightarrow \mathbb R$ be a continuous function with period $P$. Also suppose that $$\frac{1}{P}\int_0^Pf(x)dx=N.$$ Show that $$\lim_{x\to 0^+}\frac 1x\int_0^x f\left(\frac{1}{t}\right)dt=N.$$

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    That isn't true. E.g., let $f(x)=\sin(x)+1$, which has $P=2\pi$, and $N=1$, but $0\leq f(x)\leq 2$ for all $x$ implies $0\leq\int_0^xf(1/t)dt\leq 2x$ for all $x$.2012-02-02
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    @Jonas: Indeed any bounded $f$ with nonnegative $N$ is a counterexample. Perhaps OP forgot to multiply the integral by $1/x$?2012-02-02
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    I believe that the restriction on $f$ can be lightened from continuous to locally integrable.2012-02-04

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