how to show existence of limit? I'm working some problems, and the questions states; "decide if a limit exists. If it exists, find it".
But how am I supposed to do this? The tools I have at my disposal are the limit definition and what the limit is when we combine functions with known limits. 
It's a function in two variables. 
 A: The approach depends on what level of math you are at. Once you make it past calculus, mathematical rigor becomes the only acceptable way to prove results. Obtaining results through "hand-waving" is no longer acceptable. At this level of math, the limit of a sequence $(a_n)\subset \Bbb{R}$ exists if there is an $a\in \Bbb{R}$ such that given any $\varepsilon>0$, there exists an index $N \in \Bbb{N}$ such that $|a_n-a|<\varepsilon$ for all $n \geq N$. This statement is denoted as $a_n \to a$. If you are not use to rigor, the best way to think about the above statement is "as you continue on in the sequence, the difference between $a$ and any term in the rest of the sequence is as small as you want." You can equivalently show that $(a_n)$ is monotone increasing such that $a_n \leq a_{n+1} \leq a$ for all $n \in \Bbb{N}$, or $(a_n)$ is monotone decreasing such that $a \leq a_{n+1} \leq a_{n}$ for all $n \in \Bbb{N}$.
If you are beginning an introduction to limits at calculus level, you decide a limit exists if it (more or less) can be put in a form where you can "plug in" the limit without anything "bad" happening. For example, $\lim_{x\to 0} x = 0$, because we can simply plug in $x=0$ without consequence. Another example is $\lim_{y \to \infty} \frac{1}{y}=0$. This limit exists because we can always divide by any nonzero number, and the larger the number we divide by, the smaller the resulting number. These two limits are actually identical, which can be seen by the substitution $x = \frac{1}{y}$. An example of a limit that does not exist is $\lim_{x \to 0} \frac{1}{x}$, because if we plug in $x=0$ we will have divided by zero. This is bad and hence the limit does not exist. Another limit that does not exists is $\lim_{x \to \infty}e^x$. The idea here is that as $x$ gets larger, $e^x$ gets larger, and it will grow without bound. This is also bad. Since it does not stop growing, the limit does not exist. Here is an example of a problem that appears to not have a limit, but in fact it does! $$\lim_{x \to 0} \frac{\sin(x)}{x+\tan(x)}$$ If you were to plug in $x$, you'd get $\frac{0}{0}$ which is not equal to $1$. The natural reaction is to say the limit must not exist. But multiplying the above fraction by $1$ in the form of $1 = \frac{\csc(x)}{\csc(x)}$ gives us an identical fraction of $$\lim_{x \to 0} \frac{1}{\frac{x}{\sin(x)}+\frac{1}{\cos(x)}}$$ There is no problem plugging in $x=0$ in the cosine expression, as $\cos(0) = 1$. You can also prove with the sandwich theorem that $\lim_{x \to 0} \frac{x}{\sin(x)}=1$, so we can evaluate $$\lim_{x \to 0} \frac{1}{\frac{x}{\sin(x)}+\frac{1}{\cos(x)}} = \frac{1}{\frac{1}{1}+\frac{1}{1}} = \frac{1}{2}$$
This is just an introduction to limits at calculus level. If this is where you are at, there is still much to learn. You will learn to recognize limits in indeterminate form and how to solve them with L'Hospital's rule. You can extend these results and concepts to multivariable calculus as well.
