# Spivak chapter 5 problem 3 part(iv)

I am really new to analysis, and am taking an analysis 1 course. I did problem 3 part (iv) in spivak's calculus, but am really unsure if I am doing this problem's correctly. Any feedback would be greatly appreciated.

So the question asks prove the limit exists of:

$\lim\limits_{x\rightarrow a}{f(x)}=l$ where $f(x) = x^4$ and a is arbitrary.

So here's what I did

$\lim\limits_{x\rightarrow a}{f(x)}= f(a) = a^4$

Let a = 1 given ε>0 Let δ=1

$|x-1|<1$                                                             $|f(x)-l| < ε$

$0<x<2$                                                               $|x^4 - a^4|<ε$

$|x-a|< δ$                                                             $|x-a||x+a||x^2+a^2|<|x+a||x^2+a^2|δ$

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

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

$|x+a||x^2+a^2| < 15$

So

$0<|x-a|<δ=min(1,ε/15) => |f(x)-l|= |x^4-a^4| = |x−a||x+a||x^2+a^2|$

$< δ|x+a||x^2+a^2| < 15*δ = 15*(ε/15) = ε$

Someone please help me out on this, I have no idea if I answered it right or not, and I don't want to go into the test thinking I'm doing these correctly when I'm not.

By this took me 2 hours to write out because I didn't know how to use the TeX commands, so please no bullshit answers or comments.

• Here's a bullshit comment: the problem asked you to do this for arbitrary $a$ and you did it for the single specific case of $a=1$. – Randall Oct 9 '17 at 0:34
• Ok, so how would I do this for some arbitrary a? – gunnersFc Oct 9 '17 at 0:36
• plllssssssssss helpppp – gunnersFc Oct 9 '17 at 1:35

You are on the right track, you need to try to use this argument for general $a$ rather than just $a=1$. The key formula that you have found which allows you to do this is: $$|x^{4}-a^{4}|=|x-a||x+a||x^2+a^2|.$$

We want to show that this expression becomes arbitrarily small as $|x-a|$ becomes arbitrarily small. Therefore, all we really need to do is ensure that the terms $|x+a|$ and $|x^2+a^2|$ are bounded as $x\rightarrow a$. Let's make this more precise.

Observe that if $|x-a|<1$, then we have $(a-1)<x<(a+1)$. Adding $a$ shows that $(2a-1)<(x+a)<(2a+1)$.

By taking $R_{a}=\max\{|2a+1|,|2a-1|\}$, we see that if $|x-a|<1$, we have $$|x+a|<R_{a}.$$ The key point is we can bound $|x+a|$ by a positive constant $R_{a}>0$ (which will depend on $a$) provided we first bound $|x-a|$. This is similar to what you did when you showed that, for $a=1$, you had $|x+a|<|2+1|$, so you used $R_{1}=3$.

Similarly, if $|x-a|<1$, we have either $(a-1)^{2}<x^{2}<(a+1)^{2}$ or $(a+1)^{2}<x^{2}<(a-1)^{2}$ depending on the value of $a$ (specifically whether we have $-1\leqslant a\leqslant 1$ or not). In either case, we can find some constant $S_{a}>0$ such that, if $|x-a|<1$, then $$|x^{2}+a^{2}|<S_{a}.$$ Again, this is similar to what you did when you showed that, for $a=1$, you had $|x^{2}+a^{2}|<|2^{2}+1^{2}|$, so you used $S_{1}=5$.

Now we can complete the proof. Given a real number $a$ and some $\varepsilon>0$, define $R_{a}$ and $S_{a}$ as above. Now define $\delta=\delta_{a,\varepsilon}$ (that is, $\delta$ depends on $a$ and $\varepsilon$) by $$\delta=\min\left\lbrace1,\frac{\varepsilon}{R_{a}S_{a}} \right\rbrace.$$

This may seem a strange choice of $\delta$, but it is made so that $\delta\leqslant1$, and $\delta R_{a}S_{a}\leqslant\varepsilon$. You did this with $a=1$ by taking $\delta=\min\left\lbrace1,\frac{\varepsilon}{15}\right\rbrace$.

Putting this all together, if $|x-a|<\delta$, then we have \begin{align} |x^{4}-a^{4}| & = |x-a||x+a||x^2+a^2| \\ & < \delta|x+a||x^2+a^2| \\ & < \delta R_{a}S_{a} \\ & \leqslant \varepsilon. \end{align} This completes the proof.

• Thanks you! This ties up some ideas I had. Just a couple of questions I wanted to ask. When you were defining the constant $Ra$ couldn't you just let it equal $|2a+1|$? I thought taking $Ra=max|2a+1|,|2a−1|$ was a little redundant, considering $|2a+1|> |2a−1|$. Also, for clarification, the constant $Sa$ you considered the two cases of $x^2$ because you didn't know if a was positive or negative correct? Just one more thing that's bothering me, we didn't really get an explicit bound on $Sa$ like we did on $Ra$, why is that? – gunnersFc Oct 9 '17 at 3:27
• @gunnersFc Actually, $|2a+1|> |2a−1|$ holds if and only if $a>0$ (try $a=-1$ to see it fail). From this you can find $R_{a}$ explicitly for each value of $a$. In the case of $S_{a}$ you can also compute it exactly if you want. I did not do so because it again requires you to look at different values of $a$ and perform a few calculations that I felt weren't necessary to explain the proof. You can compute $S_{a}$ exactly if you wish, but the point is just that $|x^{2}+a^{2}|$ can be bounded when $|x-a|$ is bounded. – JonCC Oct 9 '17 at 3:48