Let $f(x) = 5x+9$. Show that $\lim \limits_{x \to -3}f(x)=-6$

Question

Let $f(x)=5x+9$. Show that $\lim \limits_{x \to -3}f(x)=-6$

A couple of questions about showing this and proving this. As I'm working through the problem I don't understand how I proved or showed anything as I don't understand the results I get.

I understand that $\delta$ is some range between $x$ and $L$, and $\epsilon$ is some range between $f(x)$ and $L$, so what do these proofs tell me?

An easier question, though, is while proving this, a term appears (my instructors notes) and I don't know where it came from:

We want to find $\delta$ such that $\left|f(x)-(-6)\right|=\left|(5x+9)-(-6)\right|\lt \epsilon$ whenever $0 \lt \left|x-(-3)\right| \lt \delta$

But $\left|(5x+9)-(-6)\right|=\left|5x+15\right|=5\left|x+3\right|=5\left|x-(-3)\right|\lt 5 \delta$ (where does the $5\delta$ come from?). So, we want $5\delta=\epsilon$, which implies that $\delta=\frac{\epsilon}{5}$

Answer 1

You want $\left|f(x)-(-6)\right|=5\left|x-(-3)\right|< \epsilon$, that is $\left|x-(-3)\right|\lt \epsilon/5$ and that is guaranteed if you take $\delta=\epsilon/5$.

Answer 2

Note that $|x-(-3)|<\delta\; \Rightarrow\; 5|x+3| < 5\delta = \epsilon \; \Rightarrow \; |f(x) - (-6)|<\epsilon$ as desired. We're just using a bit of foresight, obtained by doing some scratchwork.

Answer 3

The whole point of epsilon-delta proofs is given any arbitrary range around $x=-3$, you want to show that you can get arbitrary close to your supposed limit as possible.

Your range is within $\delta$ of $-3$ so:

$$0<|x-(-3)|<\delta$$

Now you just need to show that you can get arbitrary close to your supposed limit $-6$. In other words, you want to show that in your range the distance between between $f(x)$ and $L=-6$ is less than some arbitrary $\epsilon$:

So you want to show:

$$0<|x-(-3)|<\delta \implies 0<|(5x+9-(-6)|<\epsilon$$

Where

$$A \implies B$$

Means "if $A$ then $B$"

Your proof is correct, but let me advice you that in order to understand it you need to know what arbitrary means. It means "not assigned a specific value".

In terms of your other question on how we go from $|(5x+9)-(-6)|=5|x+3|$ to $|(5x+9)-(-6)|<5\delta$ ,note this just follows from your given range that $0<|x+3|<\delta$ so multiplying both sides by $5$ gives $0<5|x-(-3)|<5\delta$.

Answer 4

$5\delta$ comes from the bounds on $\delta$ you originally set up in the domain from the $\epsilon-\delta$ definition of a limit. From the original definition,you computed |f(x)-L| < $\epsilon$ and the result when the smoke cleared was 5|x -(-3)| < 5 $\delta$ since 0< |x -(-3)| < $\delta$. This supplies the choice of $\epsilon$ .

A lot of students really struggle with doing limits rigorously because they don't learn inequalities and don't realize they're really just doing what they always do in algebra with equations,just with inequalities this time.It's not different in kind from solving for x or y when working with equations.The only difference is that instead of getting a specific number, you're getting an acceptable range of values given the inequality. And as Ahmed so rightly points out in his answer, you have to get comfortable with the idea that $\epsilon$s and $\delta$s are arbitrary real numbers that satisfy the conditions in the problem.You're replacing equality with inequality, certainty with uncertainty.This is really the difference between the toy problems you learned in calculus and problems in the real world. You need to practice both and practice a lot-after awhile, you'll wonder why it was so baffling in the first place!

Answer 5

It may seem a little bit odd to be going through this rigamarole to compute the limit of $f(x)$ at $x = -3$, when one can simply evaluate $f(-3) = -6$. But the purpose of taking a limit is to find out what the function is doing around $x = -3$; this becomes important with other functions, whose actual value at a given value of $x$ (if it even exists) is not helpful.

You can think of it as a kind of game between two players.

Alice says: "$f(x)$ approaches $-6$ as $x$ approaches $-3$."

Bob says: "Naah, it just looks like it's approaching $-6$."

Alice: "Don't believe me? Then you must think that there's a value of $x$ close to $-3$ for which $f(x)$ isn't close to $-6$."

Bob: "Yeah, I guess that's what I mean."

Alice: "OK, tell you what: You pick how close you want $f(x)$ to be to $-6$. And I'll tell you how close your value of $x$ has to be to $-3$. If you can pick a value of $x$ that's close enough for me, so that $f(x)$ isn't close enough for you, you win. Deal?"

Bob: "Fine. I want $f(x)$ to be within $1/100$ of $-6$."

Alice: "No problem. Then I'm going to force you to pick a value of $x$ that's within $1/500$ of $-3$. Can you find one so that $f(x)$ is more than $1/100$ away from $-6$?"

And of course Bob cannot, because every value of $x$ in the interval $(-3.002, -2.998)$ produces a value of $f(x)$ in the interval $(-6.01, -5.99)$, just as required by Bob. The proof of the limit is a generalization of this challenge by Alice: No matter how close Bob demands $f(x)$ be to $-6$, Alice can demand $x$ to be close enough to $-3$ that $f(x)$ always satisfies Bob's demand.

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Now, why did Alice select a threshold of $1/500$? She did so specifically in response to Bob's threshold of $1/100$. Since $f(x)$ changes five times as fast, so to speak, as $x$ itself does, Alice knew she had to pick a threshold five times as tight as Bob's. That is, whatever deviation $\delta$ was permitted for $x$ would produce a deviation $5\delta$ in $f(x)$. If that had to fit within an allowance $\varepsilon$, Alice had to set $5\delta$ no larger than $\varepsilon$. She could set it smaller, if she so desired, but she had to set it no larger.

In general, when doing a proof of limits, Alice is permitted to make her threshold (the $\delta$) depend on Bob's threshold (the $\varepsilon$).