I'm self studying real analysis and currently reading about the limits of functions. Naturally everything in the chapter is about determining if a limit exists at a single point. But what about showing that a given function has limits over its entire domain? Take the class of non-rational polynomial functions. Obviously these have a limit at every $x_0$ in $\mathbb R$. My questions is, can it be proven that a given polynomial (or even the class of all non-rational polynomial functions) has a limit at every point using just what I've learned so far (formal definition of the limit of a function, the relationship between the limit of a sequence and a function, algebra of limits etc)? Or is this something that would require more knowledge?
Limits of Functions
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1@lhf. I think that CritChamp is right: he's not necessarily talking about continuity. For instance, a function like $f(x) = 0$ for every $x\neq 0$ and $f(0)=1$ is one of "his" functions because it has a limit for every point. But it's not a continuous function. – 2011-10-27
1 Answers
I think that Leslie's idea is a good one. These could be the details.
- [a] The class $S$ of functions $f$ such that $\mathrm{lim}_{x\rightarrow c} f(x)$ exists for every point $c \in \mathbb{R}$ is closed under addition and multiplication of functions.
Proof. Let $f,g \in S$, then since $\mathrm{lim}_{x\rightarrow c} (f(x) + g(x)) = \mathrm{lim}_{x\rightarrow c} f(x) + \mathrm{lim}_{x\rightarrow c} g(x) $, the limit of $f+g$ exists for every point $c$. Hence $f+g \in S$. Analogously for the product.
- [b] For all $k\in \mathbb{R}$, the constant function at $k$, $f(x) = k$ for all $x$, belongs to $S$.
Proof. Indeed, $\mathrm{lim}_{x\rightarrow c} k = k$ for all $c$.
- [c] The identity function $\mathrm{id}(x) = x$ is in $S$.
Proof. Indeed $\mathrm{lim}_{x\rightarrow c} x = c$.
From these results it follows that:
- [d] $x^n \in S$.
Proof. As we have seen [c], the result is true for $n=1$. Assume it is so for $n-1$, that is $x^{n-1} \in S$. Then $x^n = x\cdot x^{n-1} \in S$ too because of [a], [c] and induction hypothesis.
Finally,
- [e] Every polynomial is in $S$.
Proof. By induction on the degree of the polynomial. The result is true for degree zero polynomials: this is just [b]. Assume the result true for degree $n-1$ polynomials and let $p(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_{n-1} x^{n-1}+ a_n x^n$ be a degree $n$ polynomial. Then, $a_n x^n \in S$ because of [d], [b] and [a]. Hence, by induction hypothesis, $p(x)$ is a sum of two functions in $S$: $a_0 + \dots + a_{n-1}x^{n-1}$ and $a_nx^n$. Because of [a], $p(x) \in S$.
Moreover,
- [f] Every continuous function is in $S$.
Proof. If $f$ is an (everywhere) continuous function, $\mathrm{lim}_{x\rightarrow c} f(x) = f(c)$. In particular, $\mathrm{lim}_{x\rightarrow c} f(x)$ exists.
But
- [g] The class of functions $S$ is larger than the class of continuous ones.
Proof. For instance,
$ f(x) = \begin{cases} 0 & x \neq 0 \\ 1 & x = 0 \end{cases} $
is not a continuous function: $\mathrm{lim}_{x\rightarrow 0} f(x) = 0 \neq 1 = f(0)$. But $f \in S$. Indeed,
$ \mathrm{lim}_{x\rightarrow c} f(x) = 0 \ , $
for all $c$. So the limits $\mathrm{lim}_{x\rightarrow c} f(x)$ exist.
And maybe we should add a final remark:
- [h] Not every function is in $S$. :-)
For instance, $\mathrm{lim}_{x\rightarrow 0} \sin \left( \frac{1}{x} \right)$ does not exist. So the function
$ f(x) = \begin{cases} \sin \left( \frac{1}{x} \right) & x \neq 0 \\ 0 & x = 0 \end{cases} $
does not belong to $S$.