# Find the value of $\lim_ {n\rightarrow \infty} \int_{0}^{1} f( x + \frac{1}{n^2})dx$ [closed]

Define a sequence of function $$(f_n)$$ as follows $$f_n(x) = f( x+ \frac{1}{n^2})$$ where $$f: \mathbb{R} \rightarrow \mathbb{R}$$is a continious function . Find the value of $$\lim _{n \rightarrow \infty} \int_{0}^{1} f( x + \frac{1}{n^2})dx$$

My attempt : $$\lim _{n \rightarrow \infty} \int_{0}^{1} f( x + \frac{1}{n^2})dx= x^2 /2 + 0 ] _{x=0}^{x=1}= \frac{1}{2}$$

Is its true ?

• No, it's not, for instance if $f \equiv 0$. Commented Dec 21, 2019 at 22:29
• @PhoemueX how $f$ is identically equal to $0 ?$ Commented Dec 21, 2019 at 22:31

You can use a direct approach: $$\int _0^1 f(x + \frac{1}{n^2}) dx = \underbrace{\int _\frac{1}{n^2}^{1 + \frac{1}{n^2}} f(u) du}_{u-sub,\ u=x + \frac{1}{n^2}} = F\bigl(1+\dfrac{1}{n^2}\bigr) - F\bigl(\frac{1}{n^2}\bigr)$$
Now we take the limit: $$\lim_{n\to \infty} \Bigl ( F\bigl(1+\dfrac{1}{n^2}\bigr) - F\bigl(\frac{1}{n^2}\bigr) \Bigr) = \underbrace{F\bigl(\lim_{n \to \infty}(1+\dfrac{1}{n^2})\bigr) - F\bigl(\lim_{n \to \infty}(\frac{1}{n^2})\bigr)}_{because \ f \ is \ continues} = F(1) - F(0) = \int_0^1 f(x)dx$$ So a you can see, if you can show the sequence converges uniformly, you can just change the order of limit and integral, and get the same result.