Proving that $\lim\limits_{n\to\infty}\int_1^\infty\frac1{nx}dx \neq 0$ Problem:
Let $f_n(x) = \frac{1}{nx}$ for $x\in[1, \infty)$. Show that $$\lim\limits_{n\to\infty}f_n(x) = 0$$ but $$\lim\limits_{n\to\infty}\int_1^\infty f_n(x)dx \neq 0.$$
My progress:
The first limit was rather trivial given that x is positive.
For the second limit, I have come as far as to integrate $f_n(x)$, which yielded
$$\lim\limits_{n\to\infty}\left[\frac1n \left[ \lim\limits_{x\to\infty}\ln(x)\right]\right]$$
Right now, I have a $0\cdot\infty$ expression, which I suppose can be said to be not zero, but it seems cheap.
I also contemplated, since both limits tend to infinity, switch x with n, so that we have $$\lim\limits_{n\to\infty}\frac1n \cdot \lim\limits_{n\to\infty}\ln(n)$$
I tried L'Hopital on this, but that gave me 0, which is clearly not what I want.
Any help appreciated!
 A: In the improper Riemann sense you are dealing with:
$$\lim_{n \to \infty} \lim_{b \to \infty} \int_1^b \frac{1}{nx} dx = \lim_{n \to \infty} \left ( \frac{1}{n} \lim_{b \to \infty} \int_1^b \frac{1}{x} dx \right ).$$
The point of the exercise is that you cannot exchange the two limits. But the situation is simpler than you are making it, because $\int_1^b \frac{1}{x} dx = \ln(b)$. So the inner limit is always $+\infty$ regardless of the value of $n$. This means the outer limit is also $+\infty$.
What you did towards the end is a very different calculation; you showed that:
$$\lim_{n \to \infty} \frac{1}{n} \int_1^n \frac{1}{x} dx = 0.$$
This takes the limit in $b$ simultaneously with the limit in $n$.
A: Hint: $$\lim_{n\to \infty}\frac{1}{n}\underbrace{\color{#05f}{\int_{1}^{\infty} \frac{1}{x} dx}}_{\to \infty} = \infty $$
$$\color{#05f}{\int_{1}^{\infty} \frac{1}{x} dx} = \ln x \Big|_1^\infty \to \infty$$
A: $$\lim\limits_{n\to\infty}\int_1^\infty\frac1{nx}dx=\left(\lim\limits_{n\to\infty}\frac1n\right)\int_1^\infty\frac1{x}dx.$$
If the integral converges (which it doesn't), the limit is $0$. Otherwise it doesn' t exist.
