Circular Logic and Continuity So, I was doing a Calculus problem a few minutes ago and just recalled something that my real analysis professor said during a lecture years ago... 
To provide context, take the function $f$ defined by $f(x) = x+4$ for example.
Let's, for example, show that $f$ is continuous at $x = 3$.
Find $\lim\limits_{x \to 3}(x+4)$ by plugging in $3$ for $x$: you get $7$.
Since $\lim\limits_{x \to 3}f(x)= f(3)$, $f$ is continuous at $x = 3$.
Specifically, here's what I recall my professor saying:

The way continuity is taught in Calculus requires circular logic.

Clearly, I used circular logic in my example, since I assumed I could plug in $3$ to get the limit.
With a polynomial, I don't see this being too much of a problem. If I recall, there is a proof given early on in a Calculus I class which states that if $p$ is a polynomial defined by $p(x) = a_nx^{n} + a_{n-1}x^{n-1} + \cdots + a_1x +a_0$ for some positive integer $n$, then $\lim\limits_{x \to a}p(x) = p(a)$ - which I believe is proved before discussing continuity in Calculus. (We can use anything like Stewart's Calculus book as a textbook for a "typical" Calculus course.)
But how about trigonometric functions? $a^{x}$ equations for some constant $a > 0$? Natural logarithms? Powers of $x$ - $x^{b}$ - where $b$ isn't a positive integer?
How can one bypass these problems in a Calculus I course? Furthermore, is there a way to do this without using $\delta$-$\epsilon$ and just using limit theorems?
[I am willing to move this to Math Educators SE if desired, and if the question is deemed to be too broad, I can delete this.]
 A: Well, that is simply a bad proof.  What you gave does NOT prove continuity, it assumes it.  To prove continuity at x= 3 of f(x) you must prove that $\lim_{x\to 3} f(x)= f(3)$.  To do that, without the "circular reasoning" you need to show that the definition of "limit" is satisfied- and that uses the $\epsilon-\delta$ definition.
Here, f(x)= x+ 3 so f(3)= 6.  To prove that $\lim_{x\to 3} f(x)= 6$, look at $|f(x)- 6|= |x+ 3- 6|= |x- 3|< \epsilon$ and it immediately follows that you can use any $\delta< \epsilon$.  If $|x- 3|< \delta< \epsilon$ then $|x- 3|< \epsilon$ and, working backwards, $|x- 3|= |x+ 3- 6|< \epsilon$ so $|f(x)- f(3)|< \epsilon$ and we are done.
To prove this "without using δ-ϵ and just using limit theorems", use the "limit theorems"
1) f(x)= x is continuous for all x.
2) Any constant function, and, in particular, f(x)= 3, is continuous for all x.
3) If f and g are continuous at x= a then f+ g is continuous at x= a.
A: I think the saying by your professor is completely wrong. There is nothing circular here. The limit of the function $x+4$ as $x\to 3$ is not evaluated by plugging (remember that limits are not the value of the function, but they are defined in terms of values of the function).
A limit is always evaluated using theorems concerning evaluation of limits. These theorems include algebra of limits, Squeeze theorem, L'Hospital's Rule, Taylor's theorem. Apart from this we need to know certain standard limits (one for each type of elementary function). And we need two basic results which are immediate consequences of definition of limit namely $$\lim_{x\to a} k=k, \lim_{x\to a} x=a$$ Using these results and the algebra of limits we can prove that the limit of rational function at a point is equal to its value at that point, provided the rational function is defined at that point and therefore a rational function is continuous wherever it is defined. Using standard limits related to other elementary functions it is possible to prove that an elementary function is continuous wherever it is defined. 
