Here is the definition of a polynomial taken from here

A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. A polynomial in one variable (i.e., a univariate polynomial) with constant coefficients is given by $$a_0 + a_1x + a_2x^2 + \cdots + a_nx^n = \sum_{i = 0}^{n} a_ix^i$$

A polynomial is said to be an $nth$ degree polynomial if the highest power of the variable is $n$ and the leading coefficient is not equal to $0$.

Now, take the following case: $$f(x) = \lim_{a_n \to 0} \sum_{i = 0}^{n} a_ix^i = \sum_{i = 0}^{n - 1} a_ix^i$$

Now, this shows that the degree of the polynomial $$\lim_{a_n \to 0} \sum_{i = 0}^{n} a_ix^i$$

is $n - 1$. Now, here is another thing from Wikipedia

To say that $$\lim_{x \to p} f(x) = L,$$ means that $f(x)$ can be made as close as desired to $L$ by making $x$ close enough, but not equal, to $p$.

It means in the expression $$\lim_{a_n \to 0} \sum_{i = 0}^{n} a_ix^i$$

the value of $a_n$ is infinitesimally close to $0$ but $a_n \ne 0$. This means that the degree of the polynomial is $n$.

So, my question is that is it just because $a_n$ is extremely close to zero that we consider it to be equal to zero? If so, then is the degree of the expression $n - 1$?

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    $\begingroup$ A concrete example is useful. Suppose $y(x)=\epsilon x-1$: If $\epsilon$ is finite then there's a single root at $x=1/\epsilon$, but this root flies away to $\infty$ as $\epsilon\to 0$ and no other roots remain. This is an example of a singular perturbation, since the case of $\epsilon$ small but finite is not properly approximated by taking $\epsilon=0$. By contrast, when $y(x)=x-\epsilon$ then the root at $x=\epsilon$ moves to zero as $\epsilon\to 0$ and is well-behaved, corresponding to a regular perturbation. $\endgroup$ – Semiclassical Sep 16 '16 at 20:09

Your observations show that the "degree function" (which associates the degree of $p$ to each polynomial $p$) is discontinuous on the space of polynomials, in the precise sense that if $$ p(x) = \sum_{k=0}^{n} a_{k} x^{k},\quad a_{n} \neq 0, $$ then $$ \lim_{a_{n} \to 0} (\deg p) = n > \deg(\lim_{a_{n} \to 0} p). $$

  • $\begingroup$ Sorry but I still didn't understand the inequality of your last expression. $\endgroup$ – Parth Sep 17 '16 at 3:55
  • $\begingroup$ If $a_{n} \neq 0$, you have $\deg p = n$. Consequently, $\lim\limits_{a_{n} \to 0} (\deg p) = \lim\limits_{a_{n} \to 0} n = n$. $\endgroup$ – Andrew D. Hwang Sep 17 '16 at 12:46
  • $\begingroup$ But what about the inequality? I don't think that taking degree of $p$ as $a_n \to 0$ would be less than $n$. $\endgroup$ – Parth Sep 17 '16 at 12:52
  • $\begingroup$ My apology; I mis-read your comment. If $p$ has the stated form (contains no terms of degree greater than $n$) and $a_{n} = 0$ (namely, after you've taken the limit of $p$ as $a_{n} \to 0$), then $n > \deg p$. $\endgroup$ – Andrew D. Hwang Sep 17 '16 at 13:08
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Andrew D. Hwang Sep 17 '16 at 15:01

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