How to build $\varepsilon$ – $N(\varepsilon)$ proof for this limit? 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
[]My solution attempt1]1
EDIT#2:Does that work as a inequality solution?
 A: 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.
A: 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
$$\frac{\sqrt n}9{}>\varepsilon \iff n>N(\varepsilon)=\lceil 81\varepsilon^2\rceil$$
