1
$\begingroup$

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 $\infty$)?

Thanks for your time.

$\endgroup$
  • 2
    $\begingroup$ The limits of the sequences can't equal two different things. $\endgroup$ – Git Gud May 24 '14 at 16:31
  • 1
    $\begingroup$ I think that this means that the limits can be infinity or be convergent. $\endgroup$ – Valerin May 24 '14 at 16:35
  • 1
    $\begingroup$ @Jika Usually $\Bigl\{ $ denotes a conjunction, not a disjunction as it appears you want it to be. $\endgroup$ – Git Gud May 24 '14 at 16:35
  • 1
    $\begingroup$ 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 $\int_a^a f$. $\endgroup$ – Valerin May 24 '14 at 16:38
  • 2
    $\begingroup$ Maybe not. What if $u(n) = n$ and $v(n) = n^2$? $\endgroup$ – marty cohen May 24 '14 at 16:42
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.