<s>[INST] Proving limits in terms of epsilon delta questions $ f(x)= {x^3}-2x+1. $
We want to show that $ \lim_{x\to 2} ({x^3}-2x+1)= 5 $.
So here, 
$ \lvert f(x)-f(2)\rvert = \lvert {x^3}-2x+1-5\rvert$ = $ \lvert {x^3}-2x-4\rvert $. 
$ \lvert ({x^3}-2x)+(-4)\rvert \leq  \lvert {x^3}-2x\rvert + \lvert -4\rvert = \lvert {x^3}-2x\rvert + 4 $, by triangle inequality. 
I'm confused on what to do next for the $ \epsilon$ and $\delta$.  
 [/INST] As you have computed, we have 
$$|f(x)-f(2)|=|x^3-2x-4|.\tag{0}$$
Now, we have 
$$x^3-2x-4=(x-2)(x^2+2x+2)=(x-2)[(x+1)^2+1].\tag{1}$$
If $|x-2|<1$, then $1<x<3$ and we can bound 
$$0<(x+1)^2+1\leq 16+1=17.\tag{2}$$
Therefore, for any $\epsilon>0$, we can choose $\delta=\min\{ 1, \epsilon/17\}>0$
such that if $|x-2|<\delta$, then 
$$|f(x)-f(2)|=|x^3-2x-4|~~\mbox{ by }(0)\\
=|x-2|\cdot|(x+1)^2+1|~~\mbox{ by }(1)\\
\leq \frac{\epsilon}{17}\cdot 17 ~~\mbox{ by }(2)\\
=\epsilon.
$$
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