Find the limit of the converging sequence $a_n=\frac{2n+5}{3n+2}$. I have two solutions for this.
[1st method]
We claim that the limit is $\frac 23$ as $$ \left |\frac 23 -\frac{2n+5}{3n+2}\right |=\frac{11}{9n+6}< \frac{11}{9n}< \frac{2}{n}<\epsilon .$$
[2nd method]
Moreover, $$\lim_{n\rightarrow \infty} \frac{2n+5}{3n+2}=\lim_{n\rightarrow \infty} \frac{2+5/n}{3+2/n}=\lim_{n\rightarrow \infty} \frac 23$$ as $n\rightarrow \infty, 1/n \rightarrow 0$. Since the values are converging to $2/3$, the sequence will converge to $2/3$ too.
However, I am not sure about the second method. For example, how can we find the limit of the sequence $\lim (-1)^n \frac{1}{n^2}$ by the second method? Although it's clear that the sequence converges to $0$ as
$$\left |0-(-1)^n\frac{1}{n^2}\right |=\frac{1}{n^2}<\epsilon$$
 A: The second method relies on basic properties like

*

*$\lim(c + a_n) = c + \lim a_n$

*$\lim \dfrac{a_n}{b_n} = \dfrac{\lim a_n}{\lim b_n}$
provided the limits on the RHS of the equations exist.
There are many other elementary formulae for operations with convergent sequences, but that is irrelevant here.
The point in your first example is that you know that
$\lim \dfrac c n = 0$ for all $c \in \mathbb R$. But be aware that this requires a proof, although it is more or less a trivial consequence of the definition. Similarly it is easy to show that $\lim \dfrac{c}{n^k} = 0$ for all $c \in \mathbb R$ an all $k \in \mathbb N$. Let us call this type of sequences elementary zero sequences (this is just an adhoc term).
In your second example you cannot transform the sequence $a_n = \dfrac{(-1)^n}{n^2}$ into a sequence of quotients involving elementary zero sequences. So what can be done? In fact you should regard each sequence
$$a_n = \dfrac{b_n}{c_n}$$
with a bounded sequence $b_n$ and a sequence $c_n$ such that $c_n \to \infty$ as an elementary zero sequence. This is easy to prove and allows to apply the second method to many more cases. Note that $\dfrac{(-1)^n}{n^2}$ is now an elementary zero sequence in the above sense.
A: If you have a function $f(x)$ defined for all $x \geq 1$ for which $a_n = f(n)$, and $\lim_{x\to\infty}f(x)$ exists and is equal to $L$, then it is easy to see that $\lim_{n\to\infty}a_n$ exists and is equal to $L$ too. This is what the second approach is doing (CAUTION: the reverse statement does not hold, for example $a_n = \sin(\pi n)$ converges to $0$ even though $\lim_{x\to\infty}\sin(\pi x)$ does not exist).
However, a limit involving $(-1)^n$ seems like it wouldn't be able to use this method since $(-1)^x$ is not defined for all $x \geq 1$ (for example $x = 3/2$). But actually with a small modification we can analyze these in the same way. In particular, if $a_n = (-1)^nf(n)$ for a function $f(x)$ on $[1,\infty)$ such that $\lim_{x\to \infty}f(x) = L$, then $a_n$ does not converge if $L \neq 0$ and converges to zero if $L = 0$. This is because $a_n$ will oscillate between $L$ and $-L$ in the limit, so can only converge if $L = -L$, i.e. $L = 0$. So this shows that $a_n = (-1)^n\frac{1}{n^2}\to 0$ since $\lim_{x\to\infty}\frac{1}{x^2} = 0$.
