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Assignment says: Using the definition of the limit of a sequence (applying the “$\varepsilon$ – $N(\varepsilon)$ language”) prove that:

$$\lim_{n \to \infty} \frac{n+4\sqrt{n}-3}{2\sqrt{n}+7} = +\infty$$

Can you please help me?

EDIT: I solved it by separating numerator and denominator, but I don't know how to write it in Epsilon-N Form [1]My solution attempt1]1

EDIT#2:Does that work as a inequality solution?

ikerra13
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    Welcome to MSE. Please add some more context to your question. For example, you can show your attempts or the source of this assignment. – RDK Sep 26 '22 at 12:19
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    What have you tried? Where do you get stuck? – Lorago Sep 26 '22 at 12:21
  • Did you mean to write $\lim_{n \to \infty}$? – N. F. Taussig Sep 26 '22 at 12:22
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    Prove what? you have just written down the limit right now – EnEm Sep 26 '22 at 12:22
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    Do you know the definition of limit when $a_n \to \infty$? – user Sep 26 '22 at 12:25
  • it's when there exists a real number L such that for all ε>0, there exists N>0 such that. |f(x)−L|<ε for all x>N, right? – ikerra13 Sep 26 '22 at 12:27
  • Oh, now I see why I started getting stuck, something's not correct in my notes, gotta check that – ikerra13 Sep 26 '22 at 12:30
  • @ikerra13 It should be something like this: for all $\varepsilon$ there exists $N(\varepsilon)$ such that for all $n>N(\varepsilon)$ we have $a_n>\varepsilon$. – user Sep 26 '22 at 12:31
  • Let try with this and edit your question showing your work. – user Sep 26 '22 at 12:32
  • Your definition would be correct if the sequence had the limit $L\in\mathbb{R}$, however when the limit is $\infty$, the definition changes a bit to account for the fact that $\infty$ is not a real number – Lorago Sep 26 '22 at 12:32
  • OK, thank you very much! Gotta try that again with a correct definition and I'll edit a question as soon as I get some results. – ikerra13 Sep 26 '22 at 12:34
  • @ikerra13 Errata corrige to previous comment: $a_n \to L$ – user Sep 26 '22 at 12:34
  • So basically I spent time reading material a bit and solved it algebraically by separating numerator and denominator and proved that limit is infinite, but I don't know how to write it in Epsilon-N form – ikerra13 Sep 26 '22 at 13:56
  • @ikerra13 You need to solve this for $ n$ $$\frac{n+4\sqrt{n}-3}{2\sqrt{n}+7}>\varepsilon $$ and you can use some trick. – user Sep 26 '22 at 14:33
  • @ikerra13 From the given inequality you need to find $N(\varepsilon)$ such that for all $n>N(\varepsilon)$ the inequality holds. – user Sep 26 '22 at 15:48
  • @ikerra13 No this is not the way to follows. According to the definition of limit, you need to start form here $\frac{n+4\sqrt{n}-3}{2\sqrt{n}+7}>\varepsilon$ and this is ok. Now the problem is: for what value of $N$ does the inequality hold for any $n>N$? – user Sep 27 '22 at 10:28

2 Answers2

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None of the solutions you've proposed really prove it using the definition, but rather using rules of the limit. That's fine if you want to compute limits in general, and is usually how it's done, however your problem was to specifically show it using the definition. Also, they do not really seem to be correct.

I'll try to give you a somewhat detailed solution to show you how you can deal with limits which seem a bit annoying to work with, like this one.

So we want to show that

$$\lim_{n\to\infty}\frac{n+4\sqrt{n}-3}{2\sqrt{n}+7}=\infty.$$

Fix $\varepsilon>0$. We now want to find an $N$ (possibly depending on $\varepsilon$) such that

$$\frac{n+4\sqrt{n}-3}{2\sqrt{n}+7}>\varepsilon$$

whenever $n> N$ (this is what the definition says). The way we find such an $N$ is by playing around with the inequality above a bit. So let's start by just finding some smaller expression which is easier to work with and also goes to $\infty$. In particular, notice that

$$\frac{n+4\sqrt{n}-3}{2\sqrt{n}+7}\geq\frac{n-16}{2(\sqrt{n}+4)}=\frac{(\sqrt{n}+4)(\sqrt{n}-4)}{2(\sqrt{n}+4)}=\frac{\sqrt{n}-4}{2}.$$

Now clearly if

$$\frac{\sqrt{n}-4}{2}>\varepsilon,$$

then we also have that

$$\frac{n+4\sqrt{n}-3}{2\sqrt{n}+7}>\varepsilon.$$

Notice how we have just applied a few easy inequalities to reduce the problem to considering something much easier. This is a very useful technique to use when proving limits, as often it lets you simplify difficult problems a lot. Now it is easy to see that

$$\frac{\sqrt{n}-4}{2}>\varepsilon$$

if and only if

$$n>(2\varepsilon+4)^2.$$

Thus setting

$$N=(2\varepsilon+4)^2$$

we are done.

Lorago
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  • Oh my god, thank you very much!! That's really useful! – ikerra13 Sep 27 '22 at 11:51
  • I have a question tho, did we choose (n-16)/(2(sqrt(n)+4) just because we knew it is smaller then the initial inequality? – ikerra13 Sep 27 '22 at 12:07
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    We choose $\frac{n-16}{2(\sqrt{n}+4)}$ specifically because it let us do the cancellation in the next step. So we could've just as well done something different which was smaller than the original expression, but this allowed us to completely get rid of the $n$-dependence in the denominator in a really easy way – Lorago Sep 27 '22 at 12:10
  • Ooooh, okay, thank you again, this really helped a lot! – ikerra13 Sep 27 '22 at 12:12
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As an alternative, by the definition of limit we need

$$\frac{n+4\sqrt{n}-3}{2\sqrt{n}+7}>\varepsilon $$

then, in order to simplify, we can proceed as follows using that

  • $n+4\sqrt{n}-3>n \iff 4\sqrt{n}>3$
  • $9\sqrt{n}\ge2\sqrt{n}+7 \iff \sqrt{n}\ge1$

we have

$$\frac{n+4\sqrt{n}-3}{2\sqrt{n}+7}>\frac{n}{9\sqrt{n}}=\frac{\sqrt n}9{} $$

and then it suffices that

$$\frac{\sqrt n}9{}>\varepsilon \iff n>N(\varepsilon)=\lceil 81\varepsilon^2\rceil$$

user
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