Note that if $f\in L^1([0,P])$, then $ F(x)=\int_0^x(f(t)-N)\;\mathrm{d}t\tag{1} $ is a continuous function of period P.
Then, with $M=\max\limits_{[0,P]}|F|$, we have
$ \begin{align} \lim_{x\to0^+}\frac 1x\int_0^x f\left(\frac{1}{t}\right)dt &=\lim_{x\to0^+}\frac1x\int_\frac1x^\infty\frac{f(t)}{t^2}\mathrm{d}t\tag{2a}\\ &=\lim_{x\to\infty}x\int_x^\infty\frac{f(t)}{t^2}\mathrm{d}t\tag{2b}\\ &=N+\lim_{x\to\infty}x\int_x^\infty\frac{f(t)-N}{t^2}\mathrm{d}t\tag{2c}\\ &=N+\lim_{x\to\infty}x\int_x^\infty\frac{1}{t^2}\mathrm{d}F(t)\tag{2d}\\ &=N+\lim_{x\to\infty}x\left(-\frac{F(x)}{x^2}+2\int_x^\infty\frac{F(t)}{t^3}\mathrm{d}t\right)\tag{2e}\\ &=N\tag{2f} \end{align} $
$\hskip{5mm}(\mathrm{2a})$ change of variables $t\mapsto1/t$
$\hskip{5mm}(\mathrm{2b})$ change of variables $x\mapsto1/x$
$\hskip{5mm}(\mathrm{2c})$ $\displaystyle N=x\int_x^\infty\frac{N}{t^2}\mathrm{d}t$
$\hskip{5mm}(\mathrm{2d})$ $\mathrm{d}F(t)=(f(t)-N)\;\mathrm{d}t$ follows from $(1)$ and the Fundamental Theorem of Calculus
$\hskip{5mm}(\mathrm{2e})$ integration by parts
$\hskip{5mm}(\mathrm{2f})$ follows from the simple bound $\displaystyle\left|-\frac{F(x)}{x^2}+2\int_x^\infty\frac{F(t)}{t^3}\mathrm{d}t\right|\le\frac{2M}{x^2}$