Calculus – Proof of Limit Using Epsilon-Delta Definition

calculusepsilon-deltalimits

This problem was originally asked here: $\lim_{x\to 1} \sqrt{x^2 + 8} = 3$ prove this using epsilon delta. It got closed and is no longer accepting answers.

Prove this using epsilon-delta
$$\lim_{x\to 1} \sqrt{x^2 + 8} = 3$$

Definitions: $\lim_{x\to c}f(x)=L⟺∀ϵ>0,∃δ>0$ s.t. $0<|x−c|<δ⟶|f(x)−L|<ϵ.)$ as defined by Gregory Hartman et al. by the Virginia Military Institute in LibreTexts Mathematics$_1$

Context For Why This Question Is Important: Epsilon-Delta proofs of limits form the foundations of rigorous calculus as we know it. The rigor and importance of these fundamental limits should be demonstrated to students. By giving me guidance on how to correctly evaluate this limit, users provide me (and future readers of this post) with a token of knowledge.

Additional Context For the Problem: I have searched for the answer, and found this video on YouTube$_2$. https://www.youtube.com/watch?v=yC8Y50H6kw8. However, the person does not provide an example where the inequality is not easily factorable.

Given $\epsilon>0$

Choose $\delta$ such that:

Suppose $0<|x-1|<\delta$

$$|\sqrt{x^2+8}-4|<\epsilon$$
$$|\frac{x^2-8}{\sqrt{x^2+8}+4}|<\epsilon$$

The denominator is always going to be positive, so we can take it out of the absolute value brackets.

$$\frac{|{x^2-8}|}{{\sqrt{x^2+8}+4}}<\epsilon$$
[…]

Here's why I believe that this problem is interesting. Usually, epsilon-delta proofs have the convenience of being easily factorized. This one is easily solved by substituting $x=1$, but that does not respect the official definition of a limit.

Graph of Function

Here, $\epsilon=0.00001$. I know that the function's domain is the entire real numbers (the range is the nonnegative numbers) because the $x^2$ in the radicand forces the entire radicand to be positive. The square root of any positive number is well-defined in the real numbers (see below for reference).

How does one express $x^2-8$ in terms of $x-1$? Are there any other methods on how I should proceed?

Notes and Thanks:

Firstly, I would like to thank Gary for responding by telling me how I should improve this question (i.e. Show my thought process and provide context). I would also like to thank DESMOS for providing the software to produce a graphic of the function.

Citation:

  1. Hartman, Gregory, et al. “1.2: Epsilon-Delta Definition of a Limit.” Mathematics LibreTexts, December 21, 2020. https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/01%3A_Limits/1.02%3A_Epsilon-Delta_Definition_of_a_Limit. Context on the Epsilon-Delta Definition of a Limit.

  2. “Limits with Epsilon-Delta Definition! (Linear, Square Root, and Quadratic Examples).” YouTube, January 16, 2022. https://www.youtube.com/watch?v=yC8Y50H6kw8.

References:

  1. Libretexts. “3.3: Domain and Range.” Mathematics LibreTexts, October 6, 2021. https://math.libretexts.org/Bookshelves/Algebra/College_Algebra_1e_(OpenStax)/03%3A_Functions/3.03%3A_Domain_and_Range#:~:text=Another%20way%20to%20identify%20the,shown%20on%20the%20y%2Daxis.

Best Answer

Given $\epsilon>0$, instead of providing some $\delta$ out of the blue and check it "works", like textbooks too often do, we shall look for the "best" $\delta$ (i.e. the largest one which "works"). We will thus experience concretely the fact that the smaller we take $\epsilon$, the smaller $\delta$ has to be chosen.

Let us take advantage of the monotonicity of $x\mapsto\sqrt{x^2+8}$ for $x\ge0$. Assuming $|h|\le\delta\le1$, in order to get $|\sqrt{(1+h)^2+8}-3|\le\epsilon$, we just need $$3-\epsilon\le\sqrt{(1-\delta)^2+8}\quad\text{and}\quad\sqrt{(1+\delta)^2+8}\le3+\epsilon,$$ i.e., assuming moreover $3-\epsilon\ge0$: $$(3-\epsilon)^2-8\le(1-\delta)^2\quad\text{and}\quad(1+\delta)^2\le(3+\epsilon)^2-8,$$ i.e., if we even have $3-\epsilon\ge\sqrt8$: $$\delta\le\delta_0(\epsilon):=1-\sqrt{(3-\epsilon)^2-8}\quad\text{and}\quad\delta\le\delta_1(\epsilon):=\sqrt{(3+\epsilon)^2-8}-1.$$ Note that $\delta_0(\epsilon)$ and $\delta_1(\epsilon)$ are $>0$, and tend to $0$ when $\epsilon$ does.

The claim $\lim_{x\to 1} \sqrt{x^2 + 8} = 3$ is thus proved, and for every $\epsilon>0$ less than $3-\sqrt8$, the "best" $\delta$ is $$\delta(\epsilon):=\min\left(\delta_0(\epsilon),\delta_1(\epsilon)\right).$$ (We could prove that this $\min$ is $\delta_1(\epsilon)$, and that this $\delta_1(\epsilon)$ is still the "best" $\delta$ for every $\epsilon\ge3-\sqrt8$, but we don't care.)

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