# Calculate the limit with integral.

I have to calculate the limit below:

$$\lim_{n\to\infty}\int_{u(n)}^{v(n)} f(x)\mathrm{d}x.$$

I know that: $\lim\limits_{n\to\infty}u(n)=\lim\limits_{n\to\infty}v(n)\in \{a, \infty\}$, where $a\in \mathbb R$.

$f(\cdot)$ is well defined, integrable on $\mathbb{R}$.

For the first case (i.e., the limit is in $\mathbb{R}$), the limit is $0$. Is it correct to say that $\int_{a}^{a}f=0$ without any further explanation?

What about the second case (i.e., the limit is $\infty$)?

• @Jika Usually $\Bigl\{$ denotes a conjunction, not a disjunction as it appears you want it to be. – Git Gud May 24 '14 at 16:35
• The explanation is that the integral of a integrable $f$ look as a function is continuous. For this because you can write that the limit is $\int_a^a f$. – Valerin May 24 '14 at 16:38
• Maybe not. What if $u(n) = n$ and $v(n) = n^2$? – marty cohen May 24 '14 at 16:42
If the limit of $u(n)$ and $v(n)$ is the same real number $w$, for any $\epsilon>0$ $$|u(n)-w|,|v(n)-w|<\epsilon$$ holds for every $n$ big enough, hence: $$\left|\int_{u(n)}^{v(n)}f(x)dx\right| \leq |v(n)-u(n)|\cdot\sup_{x\in[u(n),v(n)]}|f(x)| \leq 2\epsilon M,$$ where $M$ is finite (given that $f$ is a Riemann-integrable function) and depends only on the behaviour of $f(x)$ near $x=w$. This clearly gives: $$\lim_{n\to+\infty}\int_{u(n)}^{v(n)}f(x)\,dx = 0.$$ In the second case, consider the function $f(x)=\frac{1}{x}$. We have: $$\int_{n}^{2n}\frac{dx}{x}=\log(2),\qquad \int_{n}^{n^2}\frac{dx}{x}=\log(n),$$ hence nothing can be said in general.