Help with $\lim_{x \to -1} x^{3}-2x=1$ I believe that I completed this problem correctly but I could use a second set of eye's to verify that I used the right methods.  Also if you have a suggestion for a better method of how to solve this I would appreciate any advice.
Prove $$\lim_{x \to -1} x^{3}-2x=1$$
Given $\epsilon > 0,$ let ; $\delta= \min\left \{{1\over2}, {4\epsilon\over11}  \right \} $
If $\; 0<\left | x+1 \right | \Rightarrow \delta \; then \; {-1\over4} < \left ( x-{1\over2} \right )^{2}-{5\over4} < {11\over4} $ 
so $\left | x^{3} - 2x - 1 \right | = \left | x+1 \right |\left | x^{2}-x-1 \right |=\left | x+1 \right |\left | \left ( x-{1\over2} \right )^{2}-{5\over4} \right |<{11\delta\over4}=\epsilon $
scratch work, let $\delta<{1\over2} \Rightarrow \left | x+1\right | < \delta < {1\over2} \Rightarrow{-1\over2} < x+1 < {1\over2}  \Rightarrow -2 < x-{1\over2} < -1 \Rightarrow 4 > \left ( x -{1\over2} \right )^{2} > 1 \Rightarrow {11\over4} > \left (  x-{1\over2}\right )^{2}-{5\over4} > {-1\over4}$

Your method looks like taken "out of the blue": why did you choose $\;\delta\,$ as you did? Why did you do that odd-looking calculations in the third line (which I didn't understand right away, btw)?.
I propose the following: for an arbitrary $\,\epsilon>0\,$:
$$|x^3-2x-1|=|(x+1)(x^2-x-1)|<\epsilon\iff |x+1|<\frac\epsilon{|x^2-x-1)|}$$
Now the estimmation "trick": for $\,x\,$ "pretty close to $\,-1\,$ , we get $\,x^2-x-1\,$ "pretty close" to $\,1\,$ (you can either use freely this or formally prove  by limits that $\,x^2-x-1\xrightarrow[x\to -1]{}1\;$).
Thus, we can choose, for example
$$\delta:=\epsilon+0.1\;,\;\;\text{so whenever}\;\;|x^2-x-1|>\frac{10\epsilon}{10\epsilon+1} $$ we'll get that
$$|x+1|<\delta\implies |(x^3-2x)-1|<\epsilon$$
