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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0Can you give an example of what you mean by a non-rational polynomial function and what you mean by a limit at a single point? – 2011-10-26
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3I think you're talking about continuity here, which should be the next chapter in the book.. – 2011-10-27
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2The theorems relating sums and products to limits immediately imply that the set $S$ of all functions $f: \mathbb{R} \to \mathbb{R}$ with the property that $\lim_{x \to c} f(x)$ exists for all $c \in \mathbb{R}$ is closed under the usual multiplication and addition of functions. So once you have proved (by hand) that constant functions are in $S$, and the function $g$ given by $g(x) = x$ is in $S$, it follows that any polynomial is in $S$. To be formal about this you might need a short proof by induction, but you don't need "more knowledge." Maybe someone will flesh this out in an answer. – 2011-10-27
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0+1 for thinking ahead of the text. @lhf is right; you're thinking about _continuity_, which is vastly important and should be next. – 2011-10-27
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0No, I'm thinking about proving that a given function (or possibly a class of functions) has a limit at each point in it's domain. It's been about 10 years since I had Calculus, but continuity might possibly be used to prove it but I'm not sure. Can the result be generalized from a single point to an entire domain. Perhaps I shouldn't have used the term 'non-rational'. What I mean is a polynomial that if not of the form p(x)/q(x). – 2011-10-27
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0@CritChamps. So, what you call "non-rational polynomial function" is just a polynomial function. And a function like $p(x)/q(x)$, with $p(x), q(x)$ polynomials is a rational function. – 2011-10-27
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2Is the question why the book focuses on proving things about limits at a single point, when in applications one often needs the existence of limits on large sets of points? The "point" (ha, ha) is that "for all $c \in D$, $\lim_{x \to c} f(x)$ exists" is a "for all" statement defined in terms of limits at a point. To prove it, you can fix a symbol $c$, assume $c \in D$, and prove that $\lim_{x \to c} f(x)$ exists. The existence of limits for large sets of $c$ is definable in terms of symbols, what is going on at a single point, and a "for all" quantifier. No new concepts are needed. – 2011-10-27
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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 $S$.