Show that $\lim_{x \to a} kx^2 + mx + n = ka^2+ ma +n$ Show that $$\lim_{x \to a}  kx^2 + mx + n = ka^2+ ma +n$$ (Assume the domain is $\mathbb{R}$) 
Let $\epsilon > 0$ be given. Suppose
$$\delta = min\{ \frac{1}{|kx + ka +m|}, 1\}$$ 
Then we have if $$ |x- a| < \delta$$ 
$$|(kx^2 + mx + n) - (ka^2 + ma + n)| = |k(x-a)(x+a)+m(x-a)| = |(x-a)(k(x+a)+m)|=|(x-a)|*|(k(x+a)+m)| < |k(x+a)+m|*\delta  = \frac{\epsilon}{|kx + ka + n|}*|kx + ka + n| = \epsilon $$
I am having some doubts about this. I chose $\delta$ to be what it is because of how |f(x)  - L| factors and because of how the largest a fraction can be is 1. It seems as though I may need to say more, in particular about whether delta can be chosen the way I have chosen it. I think I am missing something. 

By the triangle inequality we have
$$|k(x^2-a^2)+m(x-a)|\leq|k(x^2-a^2)|+|m(x-a)|=|k|\cdot|x^2-a^2|+|m|\cdot|x-a|$$
This equals the follwing, before we estimate with $\delta$
$$|k|\cdot|x-a|\cdot|x+a|+|m|\cdot|x-a|\lt |k|\delta \cdot|x+a|+|m|\delta$$
Now we use a sneaky trick to be able to estimate with with the triangle equality again
$$|k|\delta \cdot|x+a|+|m|\delta=|k|\delta \cdot|x-a+(a+a)|+|m|\delta\leq|k|\delta \cdot(|x-a|+|a+a|)+|m|\delta$$
Now we can remove all $x$ from the inequality and estimate with $\delta$
$$|k|\delta \cdot(|x-a|+|a+a|)+|m|\delta\lt |k|\delta \cdot(\delta+2|a|)+|m|\delta=|k|\delta^2+(2|ka|+|m|)\delta $$
Can you solve for $|k|\delta^2+(2|ka|+|m|)\delta-\epsilon=0$? This will give you 2 options of $\delta$, one of which will be positive. That will be your $\delta$ independent of $x$.
