finding a limit of a function by definition 
calculate the limit of the following functions and prove directly from definition (using $\epsilon$ and $\delta$)
  a) $$\lim _{x\to 2}(x^3+3x)$$

We just learned the definition of limits in regards to function and this is supposed to be a simple question. But I just can't understand how to use the info I'm given.
a) Am I right with guessing the limit is $2^3+3\cdot 2=14$?
b) Assuming I am (I don't think it would change much if I am not), then for an $\epsilon$ I want to find a $\delta$  such that for each $x$ in the punctured environment of $(x_0 - \delta,x_0+\delta)$ , $|x^3+3x-14|<\epsilon$, but I have no idea how to proceed from here. I tried searching for similar examples to see how the truth process is done but wasn't able to find any.
 A: a) You are right because $f(x)=x^3+3x$ is continuous and so $\lim_{x\to 2}f(x)=f(2)$
b) Observe that
 $$\left|x^3+3x-2^3-3\cdot 2\right|=\left|(x-2)(x^2+2x+4)+3(x-2)\right|=
\left|x-2\right|\left|x^2+2x+7\right|<\delta\left|x^2+2x+7\right|
$$
Can you proceed now?
EDIT: The term $\left|x^2+2x+7\right|$ is problematic and needs to be replaced by something containing only $\delta$. With $\left|x-2\right|<\delta$ in mind,
$$\left|x^2+2x+7\right|=\left|(x-2)^2+6x+3\right|\le \left|x-2\right|^2+6\left|x+\frac12\right|<\delta^2+6\left|x-2+\frac52\right|<\\ \delta^2+6\delta+15$$
We have
$$\delta\left|x^2+2x+7\right|<\delta^3+6\delta^2+15\delta$$
and want this to be less than $\epsilon$. If we further demand that $\delta<1$,
$$\delta\left|x^2+2x+7\right|<\delta^3+6\delta^2+15\delta<\delta+6\delta+15\delta=18\delta$$
Taking $\delta<\frac{\epsilon}{18}$ yields the result (in fact we can take
$$\delta=\frac12\min\left\{1,\frac{\epsilon}{18}\right\}$$ as our $\delta$ to be completely rigorous)
A: You can evaluate the limit by substituting $2$ because $x^3+3x$ is continuous. 
 To show that the limit is $14$, Let $\epsilon>0$. Let $\delta=\min\{1,\epsilon(\frac{1}{22})$. Now for all $x$ such that $|x-2|<\delta$, we have:
$$|x^3+3x-14|\leq|x^3-8|+|3x-6|\leq|x^2+2x+4||x-2|+3|x-2|$$
$$|x^3+3x-14|<\delta (|x^2+2x+4|+3)$$
Since for all $x\in [2-\delta,2+\delta]\subseteq [1,3]$ we know that $|x^2+2x+4|+3\leq22$
Thus:
$$|x^3+3x-14|<\delta (|x^2+2x+4|+3)<\min\{1,\epsilon(\frac{1}{22})\}(22)<\epsilon$$
