# Find the limit $\lim \limits_{n \to \infty} \int_0^1 f(nx) \,dx$

Please try to help me with a question that I'm trying to solve.

$f(x)$ is continuous in the range of $[0, \infty)$ and $\lim \limits_{x \to \infty}x^2 f(x) = 1$.

Calculate $$\lim_{n \to \infty} \int_0^1 f(nx) \,dx .$$

• Try a few examples to get a guess for the answer. – GEdgar Dec 16 '11 at 16:42
• Note that being continuous at zero matters. $f(x) = x^{-2}$ doesn't give the same result. – Michael Lugo Dec 16 '11 at 16:59

## 4 Answers

Here's an argument based on l'Hôpital's rule:

$$\begin{array}{rcl} \lim_{u \to \infty} \int_0^1 f(ux) \, dx &\stackrel{\color{Brown}{ux = y}}{=}& \lim_{u \to \infty}\frac{\int_0^u f(y) \, dy}{u} \\&\stackrel{\color{Brown}{\text{l'Hôpital}}}{=}& \lim_{u \to \infty} \frac{\frac{d}{du}\int_0^u f(y) \, dy}{\frac{d}{du}u} \\& \stackrel{\phantom{ux \to y}}{=}& \lim_{u \to \infty} f(u) \\&\stackrel{\phantom{ux \to y}}{=}& 0, \end{array}$$ since $u^2 f(u) \to 1$ as $u \to \infty$.

In this proof, we use a general version of l'Hôpital's rule that is apparently not as well-known as it deserves to be. This requires only the denominator to approach $\infty$; it is not necessary that the numerator also go to infinity. This has been explained by Bill Dubuque in several posts in this site; e.g., see here and here. The linked posts contain a formal statement of the theorem.

• Nice. It might be a good idea to point out which version of L'Hopital you're using (the "anything" over $\infty$ version; which, I don't think is very well known to students) :) – David Mitra Dec 16 '11 at 17:18
• how can you use by l'Hôpital's rule? don't you need to know that the numerator also -> to infinity? – Guy Dec 16 '11 at 17:19
• @David, Thanks for your comment. I am actually searching for the precise version right now. =) Bill Dubuque has explained this in quite a few posts; I will update my answer with a link. – Srivatsan Dec 16 '11 at 17:20
• @Guy You only need the denominator going to infinity. This is why I made my first comment. – David Mitra Dec 16 '11 at 17:20

First note that, since $f$ is continuous on $[0,\infty)$, it is integrable over any interval of finite length contained in $[0,\infty)$.

I assume you want to compute $$\lim_{n\rightarrow \infty}\int_0^1 f(nx)\,dx.$$

The above limit is equal to (making the substitution $u=nx$): $$\lim_{n\rightarrow \infty}{1\over n}\int_0^n f(u)\,du.$$

Since $\lim\limits_{x\rightarrow\infty}\bigl[ x^2 f(x)\bigr]=1$, for sufficiently large $x$, say $x\ge A$: $$\tag{1}{1\over 2x^2}\le f(x)\le {2\over x^2}$$

Now, for any number $C$ $$\tag{2}\lim_{n\rightarrow \infty}{1\over n}\int_A^n {C\over x^2}\,dx= \lim_{n\rightarrow \infty}{1\over n} \Bigl[{C\over A}-{C\over n}\,\Bigr]=0.$$ From (1) and (2), it follows that $$\lim\limits_{n\rightarrow \infty}{1\over n}\int_A^n f(u)\,du=0.$$

Thus \eqalign{ \lim_{n\rightarrow \infty}{1\over n}\int_0^n f(u)\,du &=\lim_{n\rightarrow \infty}\Bigl[\,{1\over n}\int_0^A f(u)+{1\over n}\int_A^n f(u)\,du\Bigr]\cr &=\lim_{n\rightarrow \infty} \,{1\over n}\int_0^A f(u)\ +\ \lim_{n\rightarrow \infty}{1\over n}\int_A^n f(u)\,du \cr &=0. }

$\int_0^1 f(nx)dx = \frac{1}{n}\int_0^n f(x)dx$. But we can easily show by the property of $f(x)$ that $\int_0^\infty f(x)dx = C$ exists, and that there is an $x_0$ such that $f(x)>0$ if $x>x_0$. From that we determine that, when $n>x_0$.

$$\int_0^1 f(nx)dx = \frac{1}{n} \int_0^n f(x)dx < \frac{1}{n}\int_0^\infty f(x)dx = \frac C n$$. So for $\epsilon>0$ if $n>C/\epsilon$, $\int_0^1 f(nx)dx < \epsilon$.

Now you just need t show that the limit must be non-negative, and then you've shown the limit must be zero.

We have $\int_0^1f(nx)dx=\frac 1n\int_0^nf(t)dt$ after the substitution $t=nx$. We fix $\varepsilon>0$ and $A$ such that $|x^2f(x)-1|\leq \varepsilon$ if $x\geq A$. We have \begin{align*} \left|\int_0^1f(nx)dx\right|&=\frac 1n\left|\int_0^1f(t)dt\right|+\frac 1n\left|\int_1^n\frac{t^2f(t)-1}{t^2}dt+\int_1^n\frac{dt}{t^2}\right|\\ &\leq \frac 1n\left|\int_0^1f(t)dt\right|+\frac 1n\left|\int_1^A\frac{t^2f(t)-1}{t^2}dt\right|+\frac 1n\int_A^n\frac{|t^2f(t)-1|}{t^2}dt+\frac 1n\int_1^{+\infty}\frac{dt}{t^2}\\ &\leq \frac 1n\left|\int_0^1f(t)dt\right|+\frac 1n\left|\int_1^A\frac{t^2f(t)-1}{t^2}dt\right|+\frac{\varepsilon}n\int_A^n\frac{dt}{t^2}+\frac 1n\\ &\leq \frac 1n\left(\left|\int_0^1f(t)dt\right|+\left|\int_1^A\frac{t^2f(t)-1}{t^2}dt\right|+\varepsilon+1\right). \end{align*} Now, one can find a $n_0$ such that if $n\geq n_0$ then $\left|\int_0^1f(nx)dx\right|\leq \varepsilon$, and the limit is $0$.