Calculate $\lim_{n\to\infty}\int_0^nf(\frac{x}{n})^ndx$.

For all $x\ge 0$: $$f\left(\dfrac{x}{n}\right)^n\to e^{f'(0)x}.$$

Proof:

For $x=0$, this is true since $f(0)=1$. Let $x>0$. We whish to show that $$\left|f\left(\dfrac{x}{n}\right)^n - e^{f'(0)x} \right|\to 0.$$ This is equivalent to $$\dfrac{f\left(\dfrac{x}{n}\right)^n }{e^{f'(0)x}}\to 1.$$ (Note that both denominator and enumerator are positive) Which on its turn is equivalent to $$d_n =\ln \dfrac{f\left(\dfrac{x}{n}\right)^n }{e^{f'(0)x}} \to 0.$$ We will show that this is indeed true. Note that $$d_n = n\ln f\left(\dfrac{x}{n}\right)-f'(0)x$$ and hence that (we took $x>0$) $$\dfrac{d_n}{x} = g\left(\dfrac{x}{n}\right)-f'(0).$$

We already know that $g(0) = f'(0)$, so then $d_n \to 0$.

For all $y\in[0,1]:$ $$g(y) \leq f'(0).$$

Proof:

Note that from what we know about $f$, it must be true that for all $y\in [0,1]: f(y)\leq f(0)=1$. Then it is also true that for all $y\in[0,1]:$ $$\ln f(y) \leq \ln f(0).$$ Now, it is also true that for all $y\in (0,1]:$ $$\dfrac{\ln f(y)}{y} \leq \dfrac{\ln f(0)}{y}.$$ Now, note that $\ln f(0) = 0$, so for any $\varepsilon > 0$ we get that $$\dfrac{\ln f(y)}{y} \leq \dfrac{\ln f(0)}{y-\varepsilon}.$$ Taking the limit $\varepsilon \to y$ makes the right hand side $g(0) = f'(0)$.