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This question arises from a comment made in answer to a question on proving the odd/even relationship of derivatives of even/odd functions. In particular, I am seeking clarification on the appropriateness of substitution of terms in the definition of the limit of a function.

Context: I assume the following basic definition of limit in standard real analysis. Let $A\subseteq\mathbb{R}$, $f:A\rightarrow\mathbb{R}$, and $c,L\in\mathbb{R}$. The concept of $f$ converging to $L$ at $c$ (denoted by $\lim_{x\rightarrow c}f(x)=L$) can be defined to occur when $c$ is a cluster/limit point of $A$, and when (for simplicity, here I use the basic $\varepsilon\!-\!\delta$ construct): $$ \forall\varepsilon\!>\!0,\ \exists\delta\!>\!0,\ \ (x\!\in\!A)\wedge0\!<\!|x-c|\!<\!\delta\longrightarrow|f(x)-L|\!<\!\varepsilon. $$

Question: My understanding is that the variable "x" in the prior statement (which I appreciate as written may not abide by strict rules of a formal grammar) is a bound variable (implicitly under $\forall x$), which can be substituted in with a different term so long as formal rules of substitution are followed. For example, if the limit exists, then clearly: $$ \lim_{x\rightarrow c}f(x)=\lim_{y\rightarrow c}f(y), $$ where $y$ was substituted in for the variable $x$. With respect to the comment from which this question arose, it seems (and I am simplifying here) that the following substitution ($[-x/x]$) is being made: $$ \lim_{x\rightarrow c}f(x)=\lim_{-x\rightarrow c}f(-x), $$ which translates into: $$ \forall\varepsilon\!>\!0,\ \exists\delta\!>\!0,\ \ (-x\!\in\!A)\wedge0\!<\!|-x-c|\!<\!\delta\longrightarrow|f(-x)-L|\!<\!\varepsilon. $$

If that is the case, could you substitute even more complex terms into the limit's definition? For example, would the following "limit" make sense: $$ \lim_{x^2\rightarrow c}f(x^2)? $$

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    $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – Fimpellizieri Jun 12 '17 at 3:10
  • $\begingroup$ So long as what you substitute in acts in essentially the same way. I would be cautious however in some cases, for example $\lim\limits_{x\to c} f(x)$ where $x$ ranges over all reals for the function $f(x)=\begin{cases} 1&\text{if}~x\in\Bbb Q\\ 0&\text{otherwise}\end{cases}$ clearly doesn't exist, however $\lim\limits_{q\to c}f(q)$ where $q$ only ranges over rational numbers and not all reals will actually have a defined limit of $1$. Similarly $\lim\limits_{x\to\infty} g(x)$ doesn't exist despite $\lim\limits_{\lfloor x\rfloor\to\infty} g(x)=0$ for $g(x)=x-\lfloor x\rfloor$. $\endgroup$ – JMoravitz Jun 12 '17 at 3:14
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This works, as long as you're careful about what you're saying means. What is true is that if you know that $$\lim_{x\to c}f(x)=L$$ then for any $x$, the statement $$\forall\varepsilon\!>\!0,\ \exists\delta\!>\!0,\ \ (-x\!\in\!A)\wedge0\!<\!|-x-c|\!<\!\delta\longrightarrow|f(-x)-L|\!<\!\varepsilon$$ is true. This is because, as you said, the statement in the definition of the limit is true for all $x$, and in particular it's true when you substitute $-x$ (or any other number) for $x$.

If you define $\lim_{-x\rightarrow c}f(-x)$ to refer to the number $L$ such that the statement above is true for all $x$ (if such an $L$ exists), then this means that if $$\lim_{x\rightarrow c}f(x)=L$$ then $$\lim_{-x\rightarrow c}f(-x)=L.$$ However, you should be very careful about writing something like $$\lim_{x\rightarrow c}f(x)=\lim_{-x\rightarrow c}f(-x),$$ because neither side of this "equation" is guaranteed to be defined. What you know is that if the left-hand side is defined, then the right-hand side is also defined and has the same value. The converse turns out to be true as well in this case: if the right-hand side is defined, so is the left-hand side and it has the same value. However, this is only because you can "undo" the substitution of $-x$ for $x$ (namely by substituting $-x$ for $x$ again), and this does not work for arbitrary substitutions.

In particular, it is possible that $\lim_{x^2\rightarrow c}f(x^2)$ is defined but $\lim_{x\rightarrow c}f(x)$ is not defined. For instance, if $f(x)=1$ for $x\geq 0$ and $f(x)=0$ for $x<0$, then $\lim_{x^2\rightarrow 0}f(x^2)=1$ since $x^2$ is always nonnegative, while $\lim_{x\rightarrow 0}f(x)$ is not defined.

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