Proving polynomial limit theorems I am pretty confused on this math question. It is a two-parter but I'm not sure what part a is asking me, perhaps someone on StackExchange could help.
The question reads as follows:
(a) If p is a polynomial, prove, using limit theorems, that
$$\lim_{x\to a}p\left(x\right) = p\left(a\right)$$
(b) Use the result in (a) to evaluate
$$\lim_{x\to 0^-}\left(x^3 – 8x^2 + 2x – 1\right)$$
Could somebody give me some hints about 'proving' part a with limit theorems? I was under the impression that it was a limit theorem. Because I have a polynomial wouldn't I just replace the value of $x$ with $a$? What else can I say about it? And wouldn't I just replace $x$ with $0$ in part b?

Hint: You need to use induction on degree of $p(x)$.
Update: It is a bit surprising that this was a relatively easy question on the topic of "limits" but still no good answers. I therefore expand my hint into a complete answer.
We use induction of the degree of $p(x)$. To begin induction we first prove that the result is true for polynomials of degree $0$ and degree $1$. In this process we will use the following two standard limit theorems:
1) $\displaystyle \lim_{x \to a}\{f(x) + g(x)\} = \lim_{x \to a}f(x) + \lim_{x \to a}g(x)$
2) $\displaystyle \lim_{x \to a}\{f(x) \cdot g(x)\} = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$
Let us first start with polynomials of degree $0$. In this case $p(x)$ is just a constant $k$. We have thus $p(x) = k$ for all $x$ and hence in particular $p(a) = k$. Now we need to show that $\lim_{x \to a}p(x) = p(a)$ or $\lim_{x \to a}k = k$. This is simple enough as we see that variable $x$ is not involved and thus limit of $k$ remains $k$ (this argument is informal and a proper proof is based on $\epsilon, \delta$ type argument. In fact in this case it is simple as we have $|f(x) - L| = |k - k| = 0$ which is always less than $\epsilon$ whatever $\delta$ we choose.)
Next we handle polynomials of degree $1$. Let $p(x) = Ax + B$. Then we have $p(a) = Aa + B$. We have $$\begin{aligned}\lim_{x \to a}p(x) &= \lim_{x \to a}Ax + B\\
&= \lim_{x \to a}A \cdot\lim_{x \to a}x + \lim_{x \to a}B\text{ (using limit rules }(1)\text{ and }(2))\\
&= A \lim_{x \to a}x + B\\
&= Aa + B = p(a)\end{aligned}$$ The last line uses the simple result that $\lim_{x \to a}x = a$ (again a formal proof requires that for every $\epsilon > 0$ we find a $\delta > 0$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x - a| < \delta$ where $f(x) = x$ and $L = a$. In this case $|f(x) - L| = |x - a|$ so that $\delta = \epsilon$ will do the job.)
Now we assume that the result holds for any polynomial $p(x)$ of degree $n$ so that if $p(x)$ is any polynomial of degree $n$ then $\lim_{x \to a}p(x) = p(a)$. Let us now try to see what we get if $p(x)$ is of degree $(n + 1)$. Clearly if $p(x)$ is of degree $(n + 1)$ we can write $$\begin{aligned}p(x) &= a_{0}x^{n + 1} + a_{1}x^{n} + \cdots + a_{n}x + a_{n + 1}\\
&= x\left(a_{0}x^{n} + a_{1}x^{n - 1} + \cdots + a_{n}\right) + a_{n + 1}\\
&= x\cdot q(x) + k\end{aligned}$$ where $q(x)$ is polynomial of degree $n$ and $k$ is some constant $k = a_{n + 1}$.
We then see that $p(a) = a\cdot q(a) + k$. We can now proceed in following manner $$\begin{aligned}\lim_{x \to a}p(x) &= \lim_{x \to a}\{x\cdot q(x) + k\}\\
&= \lim_{x \to a}x\cdot\lim_{x \to a}q(x) + \lim_{x \to a}k\text{ (using rules }(1) \text{ and }(2))\\
&= a\cdot q(a) + k\text{ (result holds for polynomial }q(x)\text{ of degree }n)\\
&= p(a)\end{aligned}$$ The proof by induction is now complete.
Note that the answer provided by Sami Ben Romdhane is a "technically correct statement" but it does not offer an answer to the current question from OP. What he mentions needs to be supported by a proof that polynomials are continuous and this itself is almost similar to the inductive proof I gave above (there are non-inductive proofs also, but they need some more algebra than the inductive proof I gave above).
