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 $\delta<{1\over2} \Rightarrow \left | x+1\right | < \delta < {1\over2} \Rightarrow{-1\over2} < x+1 < {1\over2}  \Rightarrow -2 < x-{1\over2} < -1 \Rightarrow 4 > \left ( x -{1\over2} \right )^{2} > 1 \Rightarrow {11\over4} > \left (  x-{1\over2}\right )^{2}-{5\over4} > {-1\over4}$

$\lim_{x \to a} f(x) = f(a)$ iff $f$ is continuous at $a$. $x^3 - 2x$ is continuous since identity and constant functions are obviously continuous and sum, difference, and product of countinuous functions is continuous. So any polynomial is continuous and any limit of polynomial is simply function value.
