Proof that a limit of $x$ goes to $a$ exists for every real $a$.

Question

I want to proof that $\lim \limits_{x \to a} 7x^2$ exists for every real a, using the epsilon-delta approach.

I know that I need to use $|x-a|<\delta$ then $|f(x)-L|<\epsilon$, but I don't have an $L$ in this situation. How would I go about solving this?

Answer 1

Let $L=7a^2$. Then $|7x^2-L|=|7x^2-7a^2|=7|x+a||x-a|$. If $|x-a|<1$, then $$|x+a|\leqslant|x-a|+2|a|<1+2|a|.$$So, given $\varepsilon>0$, take $\delta=\min\left\{1,\frac\varepsilon{7(1+2|a|)}\right\}$, and then$$|x-a|<\delta\implies\bigl|7x^2-L\bigr|<\varepsilon.$$