Question about Spivak's chapter 5 problem 3 part 3

Question

Question:

Find a $\delta$ such that $|f(x)-L|<\epsilon$ for all x satisfying $0<|x-a|<\delta$

Relating equations to the problem;

$$f(x)=x^4+\frac{1}{x} \\a=1, L=2 $$

rewriting the equation as the definition of the limits

$$|x^4+\frac{1}{x}-2|<\epsilon$$ if and only if $$0<|x-1|<\delta$$

As x approaching closer and closer to a $x^4$ approach to 1 and $\frac{1}{x}$ approach to 1 ,realizing it's the addition of limits rewriting the definition as;

$$|x^4+\frac{1}{x}-(1+1)|<\epsilon$$ now I got

$|x^4-1|<\frac{\epsilon}{2}$ if $0<|x-a|<\delta_1$ ,and I got $|\frac{1}{x}-1|<\frac{\epsilon}{2}$ if $0<|x-a|<\delta_2$

starting with $x^4$ factoring I got

$$|(x-1)(x^3+x^2(1)+1^2(x)+1^3)|<\frac{\epsilon}{2}$$

choose $\delta_1=min(1)$ , and I got $0<|x-1|<1$, using triangle inequality $|x|-|1|\le|x-1|<1$, setting $|x|-|1|<1$, and I got $|x|<2$

rewriting the expression $(x^3+x^2(1)+(1^2)(x)+1^3)\le|x^3|+|x^2||1|+|1^2||x|+|1^3|$, sub |x| in the equation I got $|2^3|+|2^2||1|+|1^2||2|+|1^3|=15$, and I just need to choose that $\delta_1= min(1, \frac{\epsilon}{30})$

for the next limit of $\frac{1}{x}$, I know I can choose $\delta_2$ such that

if $$0<|x-1|<\delta_2$$ then $$|\frac{1}{x}-1|<min(\frac{|1|}{2},\frac{\epsilon|1^2|}{4})$$

now choose $\delta=min(\delta_1,\delta_2)$ so I got $\delta=min(1,\frac{1}{2},\frac{\epsilon}{4},\frac{\epsilon}{30})$

but when I looked at the solution it is $\delta=min(1,\frac{1}{2},\frac{\epsilon}{4}, \frac{\epsilon}{8\times2\times2})$

So it is $\frac{\epsilon}{32}$ rather than $\frac{\epsilon}{30}$, could anyone check my math and see where I did wrong, and furthermore, in his final answer Spivak ignore $\frac{\epsilon}{4}$, and he got $\delta=min(\frac{1}{2},\frac{\epsilon}{32})$, can anyone give an explanation on why that is so? Is it because $\frac{\epsilon}{4}$ is bigger than $\frac{\epsilon}{32}$, and it did not matter since $\frac{\epsilon}{32}$ is closer to 1?

Answer 1

We can actually achieve a broader bound for $\delta$ through straightforward analysis. To that end, we proceed.

Note that we have

$$\bbox[5px,border:2px solid #C0A000]{x^4+\frac1x-2= (x-1)\left(x^3+x^2+x+1-\frac{1}{x}\right)}$$

Now, suppose we restrict $x$ such that $0<|x-1|<\frac12$. Then, $\frac12 <x<\frac32$. The supremum of $x^3+x^2+x+1-\frac1x$ on $(1/2,3/2)$ is $\frac{179}{24}$. Hence, $\left|x^4+\frac1x\right |< \frac{179}{24}|x-1|$

Therefore, for all $\epsilon>0$, and $\delta =\min\left(\frac12,\frac{24\epsilon}{179}\right)$,

$$\left|x^4+\frac1x-2\right |<\epsilon$$

whenever $0<|x-1|<\delta$.

Answer 2

There are many ways to solve this. We can do a bit of simplification at the beginning using the triangle inequality, and end up with simpler factors, as follows: $$\begin{aligned} \left|x^4 + \frac{1}{x} - 2\right| &= \left|(x^4 - 1) + \left(\frac{1}{x} - 1\right)\right| \\ &\leq |x^4 - 1| + \left|\frac{1}{x} - 1\right| \\ &= |x-1||x+1||x^2+1| + |x-1|\left|\frac{1}{x}\right|\\ \end{aligned}$$

Now if we constrain $|x-1| < 1/2$, we will have $|x| > 1/2$, so $|1/x| < 2$. We will also have $|x| < 3/2$, so $|x+1| \leq |x|+1 < 5/2$ and $|x^2 + 1| \leq |x^2| + 1 < 9/4 + 1 = 13/4$. Therefore we end up with $$\left|x^4 + \frac{1}{x} - 2\right| < \frac{5}{2}\cdot\frac{13}{4}|x-1| + 2|x-1| = \frac{81}{8}|x-1|$$ So if we choose $\delta = \min(1/2, 8\epsilon/81)$, then for all $|x-1| < \delta$, we will have $$\left|x^4 + \frac{1}{x} - 2\right| < \epsilon$$