Given a continuous mapping $f$ find $\lim_{n\to\infty}\int_0^1 nf(x)e^{-nx}$ Given a continuous mapping $f: \Bbb{R} \to \Bbb{R}$ find $$\lim_{n\to\infty}\int_0^1 nf(x)e^{-nx}$$ 
Now I'm clueless, however I've had to solve a similar question
$$\lim_{n\to\infty}\int_0^{\frac1{\sqrt n}}nf(x)e^{-nx}$$
I've did this by noting that since $f$ is continuous it attains it max/min on the interval $[0,\frac1{\sqrt n}]$, if we let $f(a)$ be the minimum and $f(b)$ the maximum then
$$\int_0^{\frac1{\sqrt n}}nf(a)e^{-nx}= f(a)(-e^{-\sqrt n}+1)\leq \int_0^{\frac1{\sqrt n}}nf(x)e^{-nx}\leq f(b)(-e^{-\sqrt n}+1)=\int_0^{\frac1{\sqrt n}}nf(b)e^{-nx}$$
So
$$\lim_{n\to \infty}f(a)(-e^{-\sqrt n}+1)\leq \lim_{n\to\infty}\int_0^{\frac1{\sqrt n}}nf(x)e^{-nx}\leq\lim_{n\to\infty}f(b)(-e^{-\sqrt n}+1)$$
So
$$f(a)\leq\int_0^{\frac1{\sqrt n}}nf(x)e^{-nx}\leq f(b)$$
Since $a,b\in[0,\frac1{\sqrt n}]$ and $f$ is continuous means that $f(a)=f(b)=f(0)$. However I can't do the same for the $$\lim_{n\to\infty}\int_0^1 nf(x)e^{-nx}$$ 
 A: HINT:
$$\int_0^1 ne^{-nx}f(x)\,dx=\int_0^\infty e^{-x}\xi_{x\in[0,n]}(x)f(x/n)\,dx$$
where $\xi_{x\in[0,n]}(x)$ is the indicator function.  Now apply the Dominated Convergence Theorem.
A: Hint (without Dominated Convergence Theorem). Note that by the continuity of $f$ at $0$,  for $\epsilon>0$ there is $0<\delta\leq 1$ such that $|f(x)-f(0)|\leq\epsilon$ for $x\in [0,\delta]$. Hence
$$\begin{align}
\left|\int_0^1 ne^{-nx}f(x)dx-f(0)\right|&=\left|\int_0^1 ne^{-nx}(f(x)-f(0))dx-f(0)e^{-n}\right|\\
&\leq \int_0^{\delta} ne^{-nx}|f(x)-f(0)|dx+\int_{\delta}^1 ne^{-nx}|f(x)-f(0)|dx+|f(0)|e^{-n}\\
&\leq (1-e^{-n\delta})\epsilon+(e^{-n\delta}-e^{-n})(\max_{x\in [0,1]}|f(x)|+|f(0)|)+|f(0)|e^{-n}.
\end{align}$$
Can you take it form here?
A: 
I thought it might be instructive to present another way forward that circumvents the use of the Dominated Convergence Theorem.  To that end, we now proceed.


First, we note that for any $\epsilon>0$ there exists a $\delta>0$ such that $|f(x)-f(0)|<\epsilon$ whenever $|x|<\delta$.  For a given $\epsilon>0$, fix such a $0<\delta<1$. 

Next, we write
$$\begin{align}
\int_0^1 ne^{-nx}f(x)\,dx&=\int_0^1 ne^{-nx}f(0)\,dx+\int_0^1 ne^{-nx}\left(f(x)-f(0)\right)\,dx\\\\
&=f(0)(1-e^{-n})+\int_0^1 ne^{-nx}\left(f(x)-f(0)\right)\,dx\\\\
&=f(0)(1-e^{-n})\\\\
&+\int_0^\delta ne^{-nx}\left(f(x)-f(0)\right)\,dx\\\\
&+\int_\delta^1 ne^{-nx}\left(f(x)-f(0)\right)\,dx\tag1
\end{align}$$

We can estimate the second term on the right-hand side of $(1)$ as 
$$\left|\int_0^\delta ne^{-nx}\left(f(x)-f(0)\right)\,dx\right|\le \epsilon (1-e^{-n})\tag2$$
while we can estimate the third term as
$$\left|\int_\delta^1 ne^{-nx}\left(f(x)-f(0)\right)\,dx\right|\le 2\max_{x\in [0,1]}(f(x))\,(e^{-n\delta}-1)\tag3$$

Taking the limit as $n\to \infty$ in $(2)$ and $(3)$ we find that 
$$\lim_{n\to \infty}\left|\int_0^1 ne^{-nx}(f(x)-f(0))\,dx\right|\le \epsilon$$
Since $\epsilon>0$ was arbitrary, we must have
$$\lim_{n\to \infty}\int_0^1 ne^{-nx}(f(x)-f(0))\,dx=0$$
whence we find the coveted limit
$$\lim_{n\to \infty}\int_0^1 ne^{-nx}f(x)\,dx=f(0)$$
