Prove this limit of $x^4$ using $\epsilon - \delta$

Question

$\lim_{x\to a} x^4 = a^4$ using epsilon and delta.

We know $|x^4 - a^4| < \epsilon \space \text{such that} \space |x - a| < \delta$

So we have to find a $\delta$ for, which it works.

$|x^4 - a^4| = |x^2 + a^2||x-a||x+a|$

$|x^2 + a^2||x-a||x+a| < \epsilon \space \text{such that} \space |x - a| < \delta$

Let's require,

$|x - a| < 1 \implies |x + a| < 2|a| + 1$

$1 > |x - a|$

$|x - a| \ge |x| - |a|$

Therefore,

$|x| - |a| < |x - a| < 1$

$|x| < 1 + |a|$

$x^2 < (1 + |a|)^2$

$\implies x^2 + a^2 < (1 + |a|)^2 + a^2 \implies |x^2 + a^2| < (1 + |a|)^2 + a^2$

$ \implies |x + a| < 2|a| + 1$

All together, this implies, finally:

$ |x^2 + a^2| \cdot |x+a| < ((1 + |a|)^2 + a^2)\cdot (2|a| + 1)$

This requires,

$|x - a| < \text{min}(1, \frac{1}{((1 + |a|)^2 + a^2)\cdot (2|a| + 1)})$

Thus, $\delta = \frac{1}{((1 + |a|)^2 + a^2)\cdot (2|a| + 1)})$

Thanks, please suggest if this is correct!

Answer 1

Aside from a couple of typos, it looks pretty good! Here's a cleaned up version of your proof.

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Given any $\epsilon > 0$, let $\delta = \min\left(1, \dfrac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)}\right) > 0$. Then if $0 < |x - a| < \delta$, notice that: \begin{align*} |x^4 - a^4| &= |x - a| \cdot |x + a| \cdot |x^2 + a^2| \\ &= |x - a| \cdot |(x - a) + 2a| \cdot |(x - a)^2 + 2a(x - a) + 2a^2| \\ &\leq |x - a| \cdot (|x - a| + 2|a|) \cdot (|x - a|^2 + 2|a||x - a| + 2a^2) &\text{triangle inequality}\\ &< |x - a| \cdot (1 + 2|a|) \cdot (1 + 2|a| + 2a^2) &\text{} |x - a| < \delta \leq 1 \\ &< \frac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)} \cdot (1 + 2|a|)(1 + 2|a| + 2a^2) &(\star)\\ &= \epsilon \end{align*} where the second-to-last step $(\star)$ follows because: $$ |x - a| < \delta \leq \frac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)} $$