Delta Epsilon Proof

Question

I'm having a hard time solving this delta-epsilon proof.

$$\lim_{x\to -2^+} x^4 = 16$$

Attached is my answer to the question, but apparently it is not correct and $\delta = {\epsilon \over 32}$ and not $\delta = {\epsilon\over 64}$. Can someone please explain to me what I did wrong

Answer 1

There's nothing wrong if you decide for $\delta=\varepsilon/64$ instead of $\varepsilon/32$: in these proofs it is not at all required to find “the best $\delta$”.

If the inequality $|x^4-64|$ holds whenever $0<x+2<\varepsilon/32$, then it certainly holds whenever $0<x+2<\varepsilon/64$.

It's quite difficult to follow your argument, so here's my take. We start from $x^4-64=(x+2)(x-2)(x^2+4)$.

If we impose $0<x+2<1$ (that is, we choose $\delta<1$), then $-4<x-2<-3$, hence $|x-2|<4$; also $-2<x<-1$, so $x^2<4$ and $x^2+4<8$. Therefore $$ |x^4-64|<32|x+2| $$ Hence, if we take $\delta=\min\{1,\varepsilon/32\}$, for $0<x+2<\delta$ we have $$ |x^4-64|<32(x+2)\le 32\frac{\varepsilon}{32}=\varepsilon $$ Apparently you used different bounds.

Answer 2

First note that your proof obtains the two-sided limit, not just the limit from above.

But proving more than what was requested is not an error.

For the two-sided limit, your proof is essentially correct, with one minor error.

Here's a version of your proof, with the minor error fixed, and with a few organizational changes . . .

Let $f(x) = x^4-16$.

Claim ${\displaystyle{\lim_{x \to -2}\;f(x) = 0}}$.

Fix $\epsilon > 0$, and let $\delta = \min(1,{\large{\frac{\epsilon}{65}}})$.

Our goal is to show $|x+2| < \delta$ implies $|f(x)| < \epsilon$.

Suppose $|x+2| < \delta$. \begin{align*} \text{Then}\;\;&|x+2| < \delta\\[2pt] \implies\;&|x+2| < 1\\[2pt] \implies\;&-1 < x+2 < 1\\[2pt] \implies\;&-3 < x < -1&&(*)\\[10pt] \text{Then}\;\;&-5 < x-2 < -3&&\text{[from $(*)$]}\\[2pt] \implies\;&|x-2| < 5\\[10pt] \text{and}\;\;&1 < x^2 < 9&&\text{[from $(*)$]}\\[2pt] \implies\;&5 < x^2 + 4 < 13\\[2pt] \implies\;&|x^2 + 4| < 13\\[9pt] &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Combining the results, we get}\\[9pt] &|f(x)|\\[2pt] =\;\;&|x^4-16|\\[2pt] =\;\;&|(x+2)(x-2)(x^2+4)|\\[2pt] =\;\;&|x+2||x-2||x^2+4|\\[2pt] <\;\;&(\delta)(5)(13)\\[2pt] =\;\;&(\delta)(65)\\[2pt] \le\;\;&({\small{\frac{\epsilon}{65}}})(65)\\[2pt] =\;\;&\epsilon\\[2pt] \end{align*} Thus,$\;|x+2| < \delta\;$implies $|f(x)| < \epsilon,\;$as required.

It follows that ${\displaystyle{\lim_{x \to -2}\;f(x) = 0}}$.

Comparing the solutions, here are the points which I think should be noted . . .

• At the start, state the limit to be proved.$\\[6pt]$
• Declare your choice of $\delta$ as a function of $\epsilon$, before using it.$\\[6pt]$
• You can't simply square both sides of an inequality, unless the sides are nonnegative, so in that situation, more care is required. That's your minor error.$\\[6pt]$

But all in all, your solution should definitely not be marked as wrong.

If I were grading it, I might deduct $1$ point out of $10$.