Show that $\lim_{x \to a}{P(x)} = P(a)$ for any polynomial $P(x)$

Question

In my calculus textbook I was given the following

Problem: If $P(x)$ is a polynomial, show that $\lim_{x \to a}{P(x)} = P(a)$.

I found the following solution here, were a proof was given using induction. In my textbook, the problem was given in an chapter regarding limit laws, a precise definition is only given later in the following chapter and inductions weren't used up until then. So I decided to try solve this problem using only the limit laws. In that, I am taking these for granted without proving them.

Solution: Per definition we have a polynomial $P(x) = \sum_{i=0}^n k_i x^i$ were each $k_i$ is a constant and $k, x \in \Bbb{R}$. Thus $P(a) = \sum_{i=0}^n k_i a^i$.

Now, using the limit laws for sum, multiplication and power we can write

$$ \lim_{x \to a}{P(x)} = \lim_{x \to a}{\sum_{i=0}^n k_i x^i} = \sum_{i=0}^n{\lim_{x \to a}k_i \cdot \left(\lim_{x \to a}x\right)^i} = \sum_{i=0}^n{k_i \cdot \left(\lim_{x \to a}x\right)^i}$$

and since $\lim_{x \to a}x = a$ we get

$$ \lim_{x \to a}{P(x)} = \sum_{i=0}^n{k_i \cdot \left(\lim_{x \to a}x\right)^i} = \sum_{i=0}^n{k_i a^i} = P(a) $$

$\blacksquare$

Is my reasoning correct? I am not sure, if I can use these manipulations on the sum without using induction. Thanks in advance for any comments and answers.

Answer 1

It is essentially the correct argument but personally I dislike it when these arguments are written in this way. I have typically found my point of view to be unpopular but I still try to emphasize it when I teach this stuff; let me try to explain...

Basically my issue is that in the chain of equalities, some of the earlier equalities are only justified by parts of the argument that come later. When you split the limit of a product as the product of the limits, this is only justified by the fact that both limits exist. But the fact that both limits exist is something you explain afterwards (or barely at all)!

Honestly, this is a really bad way to write down mathematics. It's one of those things that's only OK because you already know the answer i.e. it's an exercise that says "Show that....", but when you're trying to actually use mathematical rigour for real in order to find something out that you don't already know, this is the kind of thing that leads to confusion and mistakes.