Show that $ f(x)=x^4 $ is continuous at $x=-7$.
Proof: Using the $\epsilon$-$\delta$ definition I get the following:
$$|x+7|<\delta \implies |x^4-2401|<\epsilon $$
$$-\delta-7<x<\delta+7 \implies (-\epsilon+2401)^{1/4}<x<(\epsilon+2401)^{1/4} $$ ... but I'm not sure how to complete the proof from here. I know that it should be like $\delta(\epsilon) $ , but if someone could show me explicitly the next couple of steps, that would be great. Thanks
I would use the identity $x^4-a^4=(x-a)(x^3+x^2a+xa^2+a^3)$ and then say $|x^4-a^4|<|x-a|(4(|a|+1)^3)$ if $x$ is chosen in an interval that I let you find. With that you should be able to conclude.