2
$\begingroup$

I'm currently preparing for a calculus test. I was trying to solve the exercises of the test of last year, and one of the questions was:

Give a full limit research of this function: $$f:(e,\infty) \setminus \{e^{e^{k\pi}}\mid k \in \mathbb N \} \to \mathbb R $$ $$x \mapsto \frac{\cot(\ln(\ln(x)))}{\sqrt{x-e}}$$

Now, I was wondering if it is even possible that this function has a limit as $x$ approaches $+\infty$.
If I insert the function in Maple, I get $0$ as the limit for $x \to +\infty$. But I'm not really sure if that's correct since the function is periodically not defined. So my question is: Can a function that is periodically not defined have a limit for $x$ approaching infinity?

(The question wasn't formulated in English, so I'm sorry if I mistranslated it)

$\endgroup$
  • $\begingroup$ The term "periodically not defined" is unlucky. The points where $f$ is not defined are not equidistant. $\endgroup$ – Peter May 30 '15 at 9:12
  • $\begingroup$ This is a case where you want to inspect the definition of a limit Very Carefully. If the definition starts with the equivalent of, "Let $a$ be an extended real number, and let $f$ be defined in some deleted neighborhood of $a$...", then your objection is technically correct, and if you get an exam problem of this type it's worthwhile to ask the examiner for clarification. Otherwise Zev's answer gives (presumably) the intended interpretation. $\endgroup$ – Andrew D. Hwang May 30 '15 at 10:08
2
$\begingroup$

As long as a subset $S\subseteq\mathbb{R}$ contains arbitrarily large real numbers, we can say that the limit $\lim\limits_{x\to\infty}f(x)$ of a function $f:S\to\mathbb{R}$ is equal to $L\in\mathbb{R}$ if for any $\epsilon>0$, there is some $M$ for which $$x\in S,\;x> M\quad \implies \quad|f(x)-L|<\epsilon$$

$\endgroup$
  • $\begingroup$ Let me add to Zev answer this: The S can be an interval $[a,\infty)$ or maybe this interval minus a set of points $x_n$ with $\infty$ as their limit. There is only one condition on $S$; it should be unbounded from the upper side. The poorest possible set $S$ is a countable one. $\endgroup$ – Idris May 30 '15 at 9:46

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.