Prove that $\lim_{x \to a} x^4 = a^4$ using epsilon-delta definition

Question

This is an exercise in Spivak's calculus:

Let $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^4$. Prove that $\lim_{x \to a} f(x) = a^ 4$

Is my proof correct?

Proof: Let $\epsilon > 0$. If $|x-a| < 1$, it follows that $|x| < 1 + |a|$

and hence: $$|x^3 + x^2a + xa^2 + a^3| \leq |x|^3 + |x|^2|a| + |x||a|^2 + |a|^3 $$$$< ((1+|a|)^3 + (1+|a|)^2|a| + (1+|a|)|a|^2 + |a|^3$$

Choose $\delta = \min\left\{\frac{\epsilon}{(1+|a|)^3 + (1+|a|)^2|a| + (1+|a|)|a|^2 + |a|^3},1\right\}$

Then, for $x \in \mathbb{R}$ satisfying $0 < |x-a| < \delta$, we have:

$$|x^4-a^4| = |x-a||x^3 + x^2a + xa^2 + a^3| <|x-a|((1+|a|)^3 + (1+|a|)^2|a| + (1+|a|)|a|^2 + |a|^3)$$ $$< \delta((1+|a|)^3 + (1+|a|)^2|a| + (1+|a|)|a|^2 + |a|^3) \leq \epsilon$$

Hence, the result follows.

Answer 1

As mentioned in the comments, while this is correct, in such contexts, simpler bounds are generally preferred over sharp ones. So for instance, in this case, you can do

|(a+h)⁴-a⁴| = |4a³h+6a²h²+4ah³+h⁴|

From here, you can replace all the coefficients with 6. If a<1, you can replace each of its powers by 1, and if a>1, you can replace each of its powers by a³. We can assume h<1, so we can replace all of its powers by h. Hence,

|(a+h)⁴-a⁴|<|6*max(a³,1)h|

We can therefore take $\delta$ to be any number < $\epsilon$/[6*min(a³,1)]