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I'm supposed to solve this exercise: 1

Now I also have the (logically seeming) laws from the lecture slides: 2 ° 3

So at first i though you could just rewrite the term by factoring out the n^4 below the root and and then let everything inside the root except the 1 evaluate to 0 by taking the limit (and then let the n^2 cancel each other out and get 1 as result). 4

This is wrong (altough i'm not sure yet in which way/how the formulas were misapplicated). I also have the solution to the exercise: 5

My central understanding problem is in the last line of the solution: Why is it correct to let take individual limits inside the root and let everything (except the 1) evaluate to 0 in this situation, but not in the "original" formula (as i did)?

Thanks for any help


original title (if that’s more understable) (without character-limit): Limits: Why can you sometimes factor out the highest exponent under a root, and then let everything else inside the root evaluate to 0 when taking the limit, but othertimes (apparently) not?

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The rules (2 and 3) you cited have specific hypotheses, which can be easy to overlook when remembering formulae. We know that, if $a_n \to a$ and $b_n \to b$ ($a$ and $b$ finite) as $n \to \infty$, then $$a_n \pm b_n \to a \pm b.$$ Similar rules apply for multiplication, and even division if care is taken to make sure no division by $0$ happens on either side. There's also the rule for powers, as you've stated. If $a_n \to a$, with $a_n, a \ge 0$ and $r \in \mathbb{R}$, then $$a_n^r \to a^r.$$ Note that all of these rules require you to have convergent sequences to start off with.

In your working, you write

$$\lim (n^2 \cdot \sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = \lim (n^2 \cdot 1 - n^2 + 1).$$

This step, replacing $\sqrt{1 + n^{-2} + n^{-3} + n^{-4}}$ by its limit, is not supported by the above rules. If $n^2$ were also convergent (to finite $L$, say), then you could simultaneously replace it by its limit $L$, to produce the working,

$$\lim (n^2 \cdot \sqrt{1 + n^{-2} + n^{-3} + n^{-4}} - n^2 + 1) = L \cdot 1 - L + 1 = 1.$$

This would be allowed, as each sequence is convergent to some finite limit, and the rules apply. But, $n^2$ does not approach any finite limit! The limit laws do not apply. You have to use a method like the solutions, or apply L'Hopital's rule, or something else a little more clever.

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  • $\begingroup$ Thanks for the answer. (Really missed a central aspect (of the topic), so thanks a lot for the detailed explaination and for making it general (as well as relating it to the original question) (Similar rules apply for multiplication, and even division if care is taken to make sure no division by 0 happens on either side. Note that all of these rules require you to have convergent sequences to start off with. etc)). $\endgroup$
    – user631488
    Jan 4 '19 at 1:18

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