Prove that $\lim_{x \to -1^-}\frac{5}{(x+1)^3} = -\infty $ using the $\delta M$ definition of infinite limits I am posting this for you guys to let me know whether it's wrong and/or give me any advice regarding the proof.
Thank you.
Given $ M < 0 $ we need $\delta > 0$ such that $  -1 -\delta< x < -1 \Rightarrow \frac{5}{(x+1)^3} < M$
Now  $ -1 -\delta< x < -1 \Leftrightarrow  -\delta < x + 1 < 0$
Now $ \frac{5}{(x+1)^3} < M \Leftrightarrow x + 1 > \sqrt[3]{\frac{5}{M}} $
So take $\delta = -\sqrt[3]{\frac{5}{M}}$
Then  $-1 -\delta< x < -1 \Leftrightarrow -1 - (-\sqrt[3]{\frac{5}{M}})< x < -1 \Rightarrow  \frac{5}{(x+1)^3} < M$
So  $\lim_{x \to -1^-}\frac{5}{(x+1)^3} = -\infty $

Edit: The proof now starts out correctly.  There are still issues. 
For example, the line
$ \frac{5}{(x+1)^3} < M \Rightarrow$$x + 1 > \sqrt[3]{\frac{5}{M}}$
though correct, has the implication running in the wrong direction. We want to show that if $\delta$ is chosen appropriately, then $\frac{5}{(1+x)^3}$ is $\lt M$. That is not the same thing as showing that if $\frac{5}{(1+x)^3}\lt M$ then $\dots$.
Obsoleted answer below:
The argument starts out incorrectly. You need to show that for any $M$ there exists a $\delta$ such that $\dots$. 
Remark: Your handling of inequalities indicates that you can undoubtedly modify things to give a correct proof. But if the solution as it stands were being graded, it is likely that the grader would look at the first line, put an X through it, and go on to the next question. 
