One-sided exponential approximate identity (${\lim }_{n\to \infty }{\int }_0^{L}nf(x)e^{-nx}dx=f(0)$)

If $L>0$ and $f$ is bounded and continuous at zero, the positive kernel $n{e}^{-nx}$ 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.