Prove that the limit, $\displaystyle\lim_{x \to 1^+} \frac{x-3}{3 - x - 2x^2}$, exists. For each of the following, use definitions (rather than limit theorems) to prove that the limit exists. Identify the limit in each case.
d)$\displaystyle\lim_{x \to  1^+}  \frac{x-3}{3 - x - 2x^2}$
Proof: We need to prove the limit exists as $x$ approaches $1$ from the right. 
Let $M$ be a real number. Then we need to show $f(x) > M$, without loss of generality, suppose $M > 0$. Then as $x \to 1^+$, $x-3 \to -2$ and observe the denominator approaches $x \to 0$ as $x \to 1^+$. Also note the parabola has roots $\frac{-1}{2}$ and $1$. Therefore, let $0 < \delta < 1$ such that$ 1 < x < 1 + \delta$
thus, $0 < 3 - x - 2x^2 < \frac{1}{M}$
That is $\frac{1}{3 - x - 2x^2} > M > 0$ then $\frac{x-3}{3 - x - 2x^2} < M$  for all $1 < x < 1 + \delta$
Can someone please help me finish? If it needs work, please help  me. Thank you very much.

To prove that the limit doesn't exist, it suffices to show that for any $M > 0$, there exists a $\delta$ such that for $0 < x - 1 < \delta$, $|f(x)| > M$. (Then, $f$ is unbounded as $x \rightarrow 1^{+}$.) One way to reverse engineer a proof is find an expression in terms of $\delta$ that will bound $|f(x)|$ as $\delta \rightarrow 0$; using that expression, you can then set $M$ equal to your bound, and deduce what $\delta$ would have to be so that for $0 < x - 1 < \delta$, $|f(x)| > M$. In this process, it is convenient to make certain simplifying assumptions (without loss of generality) to make deducing a bound easier. For instance, you might find it helpful to deduce a bounding expression in terms of $\delta$ that holds when $0 < \delta < 1/4$. (Or, instead of 1/4, pick a convenient upper bound that helps you derive an upper bound; I found 1/4 was convenient for this purpose.)
