I am currently working my way through Spivak’s Calculus, and I cannot figure out the last basic algebra step to reduce the problem so that it reads as presented in the book’s solutions.

Specifically, I am asked to **find a $delta$ such that $lvert{x^4 – a^4}rvert < epsilon ; forall x$ satisfying $lvert{x-a}rvert < delta$.**

The way to tackle the problem is very straightforward. Notice that $lvert{x^2 – a^2}rvert < min{big(frac{epsilon}{2(lvert{a^2}rvert + 1)}, 1big)} implies lvert{x^4 – a^4}rvert < epsilon$. Furthermore $lvert{x – a}rvert < min{big(frac{epsilon’}{2(lvert{a}rvert + 1)}, 1big)} implies lvert{x^2 – a^2}rvert < epsilon’$. Substituting $epsilon = epsilon’$, we have:

$$

min{bigg(frac{min{big(frac{epsilon}{2(lvert{a^2}rvert + 1)}, 1big)}}{2(lvert{a}rvert + 1)}, 1bigg)}

$$.

After I simplify, I choose:

$$

delta = min{bigg(min{big(frac{epsilon}{4(lvert{a^2}rvert + 1)(lvert{a}rvert + 1)}, frac{1}{2(lvert{a}rvert + 1)}big)}, 1bigg)}

$$

**My Question**:

Spivak’s solutions show that this may be reduced to

$$

delta = min{big(frac{epsilon}{4(lvert{a^2}rvert + 1)(lvert{a}rvert + 1)}, 1big)}

$$

and I cannot figure out why this is true. Why can we guarantee that

$$

begin{align}

frac{epsilon}{4(lvert{a^2}rvert + 1)(lvert{a}rvert + 1)} &< frac{1}{2(lvert{a}rvert + 1)} \

frac{epsilon}{2(lvert{a^2}rvert + 1)} &< 1

end{align}

$$

for any such $epsilon$?