I need to find good examples as to how prove that the sequence converges to a particular limit using delta-epsilon definition of the limit. So, thus far I only have seen online very easy examples where upon simplification you get a rational expression like $\frac{1}{n+1}$ or $\frac{2}{n}$ or if there are more complicated cases they are not proved using the definition. So, my question is, do anybody know some good online source where the ways to compare the expression to that of smaller or greater one leads to conclusion that the sequence converges. For example if I have to prove that $$\left|\frac{n^3+2n^2+3}{4n+3}-2\right|<\epsilon$$ How do I know to what expression can I compare it to conclude that the sequence converges? I know that generally if the sequence is said to converge to some finite value then to prove that it converges you have to fins some larger expression then solve it for n, and then for any N greater than n the equality will hold. So for example in this instance we have to prove that:$$\left|\frac{n^3+2n^2+3-8n-6}{4n+3}\right|=\left|\frac{n^3+2n^2-8n-3}{4n+3}\right|<\epsilon$$. We know that $4n+3>n$ and $n^3+2n^2-8n-3<n^3+2n^2$ so we get that:$$\left|\frac{n^3+2n^2-8n-3}{4n+3}\right|<\left|\frac{n^3+2n^2}{n}\right|=\frac{n(n^2+2n)}{n}=n^2+2n$$. Now, it most be proved that there exists such $N$ that any $n\in N$ greater or equal to $N$ satisfies the inequality $$n^2+2n<\epsilon$$. But I have no idea how to express n in relation to epsilon, how should I proceed? So, it is known that the limit is not 2, how to prove it?
If I must prove that $$\frac{n^3+2n^2+3}{4n+3}>\frac{1}{\epsilon}$$. So $n^3+2n+3>1 \forall n\in N$ and $4n+3<5n \ \ \forall n\in N$ then I must prove that $$\frac{1}{5n}>\frac{1}{\epsilon}.$$ Or in other words $5n<\epsilon \Rightarrow n<\frac{\epsilon}{5}$. So I am now stuck with not being able to find $N$ such that all $n\ge N \ \ n\in N \ \ \frac{n^3+2n^2+3}{4n+3}>\frac{1}{\epsilon}$
And my initial question still applies, if you know any good source where it is explain how to use algebraic manipulations to prove the limit, I would be very glad, if you share it!
For rational functions of $n$ such as this, it is very easy to figure out and prove their limits. Just divide every term, top and bottom, by the highest power of $n$ in the problem:
$$\frac{n^3+2n^2+3}{4n+3}= \frac{\dfrac{n^3}{n^3} + \dfrac{2n^2}{n^3} + \dfrac{3}{n^3}}{\dfrac{4n}{n^3} + \dfrac{3}{n^3}} = \frac{1 + 2\dfrac{1}{n} + 3\dfrac{1}{n^3}}{4\dfrac{1}{n^2} + 3\dfrac{1}{n^3}}$$
Now, as $n \to \infty$ any term with $n$ in the denominator (to any power) will go to $0$: $$\frac{1 + 2\dfrac{1}{n} + 3\dfrac{1}{n^3}}{4\dfrac{1}{n^2} + 3\dfrac{1}{n^3}} \to \frac{1 + 2\cdot 0 + 3\cdot 0}{4\cdot 0 + 3\cdot 0} = \frac 10 = \infty$$ (where the last "=" should be taken as an abbreviation for an argument that can be made, not a definitive equality).
Because you are dividing by the highest power in your rational function, there will always be at least one term that does not go to $0$. So you will always end up with one of these forms, with $A,B \ne 0$:
$$\frac A0 = \pm\infty, \quad \frac 0B = 0, \quad \frac AB$$
The first occurs when the highest degree term occurs in the numerator, the second when it occurs in the denominator, and the third when it occurs in both.
For sequences with more general functions of $n$, the situation is harder. A similar trick may work, if you have a ratio with some term dominating. For example, in $$\frac{ 2^n + n^2 - 1}{7\cdot 2^n - n^{27}}$$ $2^n$ grows faster than the other terms, so if we divide through by it: $$\frac{ 2^n + n^2 - 1}{7\cdot 2^n - n^{27}}= \frac{ 1 + \dfrac{n^2}{2^n} - \dfrac 1{2^n}}{7 - \dfrac{n^{27}}{2^n}}\to \frac {1 + 0 - 0}{7 - 0} = \frac 17$$
But there are plenty of functions where this cannot be done, forcing a more thurough examination. There isn't a single trick that always works. An example is the sequence $$\sin n$$ which neither converges nor diverges to $\pm \infty$, but keeps moving randomly back and forth across the interval $[-1,1]$. But it takes some intimate study of the unit circle to show this.
I haven't touched on $\epsilon\text{-}\delta$ in the above. Since we are talking about sequences, what we actually need is $\epsilon\text{-}N$, not $\delta$.
The harder part of any limit proof is figuring out what the sequence is doing. Once you've figured out what the limiting behavior is, the process of creating a proof becomes fairly straight forward (it can still be a pain to figure out what $N$ will work for each $\epsilon$, but you know where to look to figure it out.
In your original example, we need to prove divergence to $\infty$, so $\epsilon$ also becomes $M$. That is, we need to show that for each $M$, there is an $N$ such that if $n > N$, then $f(n) > M$.
But we know that for any $n \ge 1$, $1 + 2\frac{1}{n} + 3\frac{1}{n^3} > 1$ and $4\frac{1}{n^2} + 3\frac{1}{n^3} \le \frac 7n$ for any $n > 1$, and so $$f(n) > \frac 1{4\dfrac{1}{n^2} + 3\dfrac{1}{n^3}} > \dfrac 1{\dfrac 7n} = \frac n7$$ So if we take $N = 7M$, we have $$n > N \implies f(n) > \frac n7 > \frac N7 = \frac{7M}7 = M$$ as required.