Finding, not verifying, the limit of a sequence. If I'm told to prove that $\lim \limits_{n \to \infty}{\frac{1}{n^2}} = 0$, I would do the following:
$|\frac{1}{n^2} - 0| = \frac{1}{n^2} < \epsilon \implies N(\epsilon) = \frac{1}{\sqrt{\epsilon}} < n$
Every $\epsilon > 0$ is defined, so $\frac{1}{n^2} \to \infty$.

However, what if I'm not told $L = 0$? How would I find that $L = 0$, without looking at graphs or testing increasingly larger $n$?
I used $a_n = \frac{1}{n^2}$ as a simple example. I'd prefer some method that works for other sequences, too.
 A: Finding/evaluating a limit requires the use of limit theorems like


*

*algebra of limits

*Squeeze theorem 

*Cesaro Stolz theorem

*standard limits like $1/n \to 0$ as $n\to\infty$ which are proved using definition of limits.


Apart from these techniques sometimes we need to use theorems on existence of limits like a monotone and bounded sequence tends to a limit and then use the this fact to find the limit. Such solutions are often tricky. I show one example. 
Let $a_n=x^{n}$ where $x$ is real. I deal with  the case when $0<x<1$. Clearly $a_n$ is monotone and bounded and hence it tends to some limit $L$.  Now we know that $a_{n+1}=xa_{n}$ and taking limits as $n\to \infty$ we see that $L=xL$ and hence $L=0$.
Your question about sequence $1/n^{2}$ can be solved by using product rule and the standard limit $1/n\to 0$. In fact we can generalize this and use the result $1/n^{a}\to 0$ for $a>0$ as a standard limit. Sometimes a combination of the techniques mentioned above may be required.
The above techniques exclusively deal with theorems related to sequences. Sometimes we also need results which deal with limits of functions of a real variable. Two such results which are popular are as follows:


*

*If $a_n\to L$ and a function $f$ is continuous at $L$ then $b_n=f(a_n) \to f(L) $.

*If $f(x) \to L$ as $x\to\infty$ then $f(n) \to L$ and if $g(x) \to L$ as $x\to 0$ then $g(1/n)\to L$.
Thus for example $n\sin(1/n)\to 1$ as $n\to\infty$. 
A: Prove that $1/n^2$ cannot converge to $L$ when $L\ne 0.$ As follows: 
(1).  If $L<0$ and $0 <\epsilon <|L|/2$ then $|L-1/n^2|>\epsilon$ for ALL $n$.
(2). If $L>0$ there exists $n_0\in \mathbb N$ with $n_0>\sqrt {2/L}.$ Then with $\epsilon  = 2/L$ we have :  $\quad |L-1/n^2|>\epsilon$ for all $n\geq n_0$.
