For example:
If you wanted to prove $\displaystyle \lim_{x \to 3} x^{2}+2x = 15$
you could simplify it to: $\mid x+5 \mid \mid x-3 \mid < \varepsilon$
and to solve you can choose $\delta \leq 1$ and so on, but does it matter what number delta is less than? Could you choose $\delta \leq 100$ and get a correct answer, like $\min{(100,\frac{\varepsilon}{105})}$? Is there an example (more complicated/involved than this) where delta must be less than or equal to a specific value?
Also, why can't we simply choose $\delta = \frac{\varepsilon}{\mid x+5 \mid}$? Why does it matter if there is an x in the expression for delta? Does this cause any problems in certain functions or something?
You are mischaracterizing what the $\epsilon$-$\delta$ definition of the limit of a function is. We say $$\lim_{x\to{p}}f(x)=L$$ if and only if $$\forall{\epsilon\gt0},\exists{\delta\gt0},\forall{x\in\mathrm{dom}(f)},\left[0\lt|x-p|\lt\delta\implies|f(x)-L|\lt\epsilon\right].$$
As such, the statement $$\lim_{x\to3}x^2+2x=15$$ cannot merely be simplified to $$|x^2+2x-15|\lt\epsilon$$ as you say. You need the full definition in there. The full definition also explains why $\delta$ cannot depend on $x$: just look at the quantifiers in the definition. Since the given condition must hold true for all $x$, $x$ cannot be a fixed, arbitrary quantity that one can freely vary to adjust $\delta$. To state it more succinctly: $x$ is a bound variable, not a free variable.
So how does one prove that $$\forall{\epsilon\gt0},\exists{\delta\gt0},\forall{x\in\mathbb{R}},\left[0\lt|x-3|\lt\delta\implies|x^2+2x-15|\lt\epsilon\right]?$$ Well, notice that $|x^2+2x-15|=|(x+5)(x-3)|=|x+5||x-3|=|(x-3)+8||x-3|$. By the triangle inequality, $|(x-3)+8|\leq|x-3|+|8|=|x-3|+8$, hence $|(x-3)+8||x-3|\leq|x-3|^2+8|x-3|$. As such, $$\forall{x\in\mathbb{R}},\left[0\lt|x-3|\lt\delta\implies|x^2+2x-15|\leq|x-3|^2+8|x-3|\lt\delta^2+8\delta\right].$$ We would like for $|x^2+2x-15|\lt\delta^2+8\delta$ to be equivalent to $|x^2+2x-15|\lt\epsilon$, so that we can complete our proof. This could be accomplished by having $\epsilon=\delta^2+8\delta$. What we need to prove, thus, is that for $$\forall{\epsilon\gt0},\exists{\delta\gt0},\epsilon=\delta^2+8\delta.$$ The equation $\epsilon=\delta^2+8\delta$ is equivalent to $\epsilon+16=\delta^2+8\delta+16=(\delta+4)^2$. Hence $\sqrt{\epsilon+16}=|\delta+4|$. Since we require $\delta\gt0$, it follows that $|\delta+4|=\delta+4$, thus $\sqrt{\epsilon+16}=\delta+4$, equivalent to $\delta=\sqrt{\epsilon+16}-4$. This will allow you to finish the proof.