= One-sided exponential approximate identity
{title2=$\lim_{n\to\infty}\int_0^Lnf(x)e^{-nx}\,dx=f(0)$}
If $L>0$ and $f$ is bounded and continuous at zero, the positive kernel $ne^{-nx}$ has mass $1-e^{-nL}$ on $[0,L]$, tending to one, and mass at most $e^{-n\delta}$ outside $[0,\delta]$. Split the integral of $f(x)-f(0)$ at $\delta$: <continuity> makes the near part arbitrarily small, and boundedness times the exponentially small tail controls the far part. This proves the displayed endpoint <approximate identity>. A shrinking cutoff $L_n=n^{-1/2}$ still captures mass $1-e^{-\sqrt n}$ and therefore gives the same limit.
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