Question: Thank you for taking the time to read my question.

The scenario is, we're playing a game where we have in a box 9 balls: 4 red, 3 green, 2 blue.

We draw one ball at a time, and note down its color and then return it for the next draw, the game stops only when we have drawn the same color twice in a row.

* What is the probability of the game ending after $n$ draws? $P(X=n)$?
* What is the probability of the last two balls being blue?

I have tried to treat it as a geometric probability distribution, with: p: probability of having two balls in a row with the same color $P(X=n) = n-1$ fails and then a success but this ball number 2 and 3 could be the same color in this. I've tried to discuss it case by case, starting with if the first ball is Blue, Red, Green, in hopes of noticing some recursion, but no luck. Haven't had any other ideas.

Answer: 
The definition for limit of a function at a point is as follows:

If for all $\epsilon>0$, there exists $\delta>0$ such that $0<|x-x\_0|<\delta$ implies $|f(x)-L|<\epsilon$, then the limit of $f(x)$ as $x$ approaches $x\_0$ exists and is equal to $L$.

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Here is a correct version of the $\epsilon,\delta$ argument using the definition:

Let $\epsilon>0$ be given and define $\delta=\min\left\{\epsilon,\frac{1}{2}\right\}$. Then $|x-2|<\delta$ implies $\frac{3}{2}<x<\frac{5}{2}$. However, we also have that

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|=\frac{|4-x^2|}{4x^2}=\frac{|2-x||2+x|}{4x^2}=\frac{|x-2|(x+2)}{4x^2}$$

and therefore

$$<\frac{\delta(5/2+2)}{4\cdot (3/2)^2}=\frac{1}{2}\delta$$

Now, if $\delta=\epsilon$ then we are done as

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|<\frac{1}{2}\delta=\frac{1}{2}\epsilon<\epsilon$$

If not, then $\delta=\frac{1}{2}<\epsilon$ and therefore

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|<\frac{1}{2}\delta=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}<\frac{1}{2}<\epsilon$$

and we are done.
You seem to be incorporating some version of an $\epsilon$-$\delta$ argument, which is unnecessary if you're using basic limit theorems. For the record though, your statement
"Given that when $|x-2|<\delta$ then $\frac{1}{x^2}=\frac{1}{4}$" is a misrepresentation of $\epsilon$-$\delta$ arguments. It would be $|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$, as you have later down. With regards to that portion, though
"I proceeded with: $\lim\_{x\to 1}\frac{1}{x^2}=\frac{1}{4}$, $|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$" is a bit of a reversal of the chain of reasoning you would present in an $\epsilon$-$\delta$ argument.

Anyhow, I just wanted to point those things out, but since this is not supposed to be an $\epsilon$-$\delta$ argument, I'll leave it there. As for the question at hand, Correa is correct in his comment.

Let $f(x)=1$ and let $g(x)=x$. You can use the property you stated, namely $\lim\_{x\to p}\frac{f(x)}{g(x)}=\frac{f(p)}{g(p)}$ (provided each limit exists separately and $g(p)\neq 0$), but you must first establish that $\lim\_{x\to 2}x=2$. You can then use $\lim\_{x\to p}f(x)g(x)=\lim\_{x\to p}f(x)\lim\_{x\to p}g(x)$ to get the final answer.