Prove that $\lim\limits_{x\to 2} (3x + 4) = 10$.

Question

Given this absolutely simple limit and given that it is already known how to prove it using the $\epsilon-\delta$ definition, I am wondering the following: What is the difference between proving this limit the rigorous way (i.e., $\epsilon-\delta$) vs. "proving" computationally (i.e., plugging in x = 2)? A better question, what more information does proving it the rigorous way provide than computationally?

Here is a setting to guide the reader for one answer of the many possible ones that could be given:

A friend who is taking the Calculus I (the first standard course in the undergrad calculus sequence in college) is with me doing hw or something else. I ask him to prove the given limit above. He plugs in $x = 2$ and says "done". Is he wrong? If so, then why? If not, then why not? If in between, explain also?

NOTE: I am not asking how to prove this. However, if proving this helps in the explanation of my question, then go ahead.

Answer 1

All polynomials are continuous functions over the real numbers, by definition this means that at each point the limit exists and is equal to the value of the function. The $\epsilon,\delta$ is not necessarily meant to provide clarity but rather is the foundation of how limits work.

Answer 2

It's not always true actually that $$ \lim_{x\to a}f(x) = f(a) $$ functions that do satisfy this property, however, are called "continuous functions". Not all functions are continuous, you can find many many examples, perhaps the simplest are piecewise functions: $$ f(x) = \left\{\begin{array}{cc}0&x \in [0,1)\\1&x=1\end{array}\right. $$ notice here, that ${\lim_{x\to 1}f(x) = 0}$ but ${f(1)=1}$.

So just plugging in ${x=2}$ for the limit is only valid if you can say that ${3x+4}$ is continuous. This is indeed true, but you must prove it's continuous, or use some theorem that allows you to say it's continuous. In fact, all polynomials will be continuous. The ${\epsilon-\delta}$ proof is required if you do not prove/know continuity.

Answer 3

In rebuttal to the answer of guy3141, it could be argued that the assumption that $f(x) = (3x + 4)$ is a continuous function, which it certainly is, represents circular reasoning. It depends on the intent behind the question.

If you are allowed to use theorems that pertain to continuity, and the fact that $f(x)$ is a continuous function, then I agree with the answer of guy3141.

If you are not allowed to assume continuity, then consider that the normal definition of a limit is that

for each $\epsilon > 0$ there exists $\delta > 0$ such that
$0 < |x -2 | < \delta \implies |f(x) - 10| < \epsilon.$

Note that under the $\epsilon, \delta$ scenario, any consideration of the value of $f(x)$ at the particular value of $(x = 2)$ is out of bounds.

This is because of the constraint $0 < |x - 2| < \delta$
as opposed to the constraint $0 \leq |x - 2| < \delta.$