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Suppose we have two functions $f,g:\mathbb{R}\to{\mathbb{R}}$ such that:

$$\lim_{x\to{}\alpha}{f(x)}=L \text{ and }\lim_{y\to{\beta}}{g(y)}=\alpha.$$

Is then true that:

$$\lim_{y\to{}\beta}{f(g(y))=L}?$$

My intuition tells me yes. Using an $\epsilon-\delta$ approach seems very long and tedious. Is that the case? Or is there a nice way of showing this to be true. Or, is there a nice counterexample?

Kenta S
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kam
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3 Answers3

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Here is a counterexample :

$$g(x)=1\;if\; x\ne 0\; and\;g(0)=2$$

$$f(x)=3\;if\;x\ne 1\; and\;f(1)=4$$

thus

$$f(g(x))=4\;if\;x\ne 0\;and\;f(g(0))=3$$ take $$\alpha=1\;and \;\beta=0$$

we have $$\lim_{x\to 1}f(x)=3=L$$ $$\lim_{x\to 0}g(x)=1$$ but $$\lim_{x\to 0}f(g(x))=4\ne 3$$

hamam_Abdallah
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  • This relies on what is, as far as I know, a less common definition of a limit based on punctured neighborhoods - and strikes me as very misleading because of it. This answer at least needs to make the used definition of a limit being used very prominent and to recognize that the answer requires that exact definition. Without this qualification, I think the answer does more harm than good. – Milo Brandt Jan 08 '21 at 21:45
  • To add to my last comment: I looked through the textbooks on analysis I have readily accessible - all three use the definition of a limit based on neighborhoods and this answer is incorrect in that context (although one only discusses this kind of limit with respect to metric spaces, not $\mathbb R$ specifically). I checked Wikipedia, where the definition would allow this (citing Spivak's Calculus). It might be a definition that gets used in calculus, but largely not used in analysis and topology. – Milo Brandt Jan 08 '21 at 22:03
  • @MiloBrandt The definition based on punctured neighborhoods is pretty standard in analysis. For example, Rudin's analysis textbook uses it. Yes, there are some texts which do not use it. This is often how things are in calculus / analysis: the same concept is overloaded with different definitions. But I wouldn't call this one less common. – bjorn93 Jan 09 '21 at 03:48
  • @bjorn93 I could believe that - I guess my concern is not about the relative frequencies of the definitions, but more that the neighborhood one is common enough that it's likely that someone using the neighborhood definition could happen across this question and see two downvoted answers (one of which does mention the ambiguity) and one upvoted answer (which does not) and then be seriously misled on an important property of of the kind of limits they're working with. This can and should be corrected by incorporating the definition of a limit into this answer. – Milo Brandt Jan 09 '21 at 05:00
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Suppose $\alpha\notin g(\mathbb R\setminus\{\beta\})$.

Let $\epsilon>0$ be arbitrary. Then there exists a $\delta>0$ such that $0<|x-\alpha|<\delta\Rightarrow|f(x)-L|<\epsilon$.

Furthermore, there exists a $\eta>0$ such that $0<|y-\beta|<\eta\Rightarrow|g(y)-\alpha|<\delta$. Since $g(y)\ne\alpha$ by our hypothesis, we have $0<|g(y)-\alpha|<\delta$. Then, by my statement above, this implies that $|f(g(y))-L|<\epsilon$, which shows that $\lim_{y\to\beta}f(g(y))=L$.

P.S.: You can get away with assuming that for a small enough $\epsilon>0$ there are no $x\in\mathbb R\setminus\{\beta\}$ with $|x-\beta|<\epsilon$ such that $\alpha=g(x)$.

When this is not the case, there is a sequence of points $y_n\in\mathbb R\setminus\{\beta\}$ with $y_n\to\beta \ (n\to\infty)$ such that $g(y_n)=\alpha$. Then, $\lim_{n\to\infty}f(g(y_n))=f(\alpha)$, so that $\lim_{y\to\beta}f(g(y))=f(\alpha)$ if this limit exists.

Kenta S
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  • Thank you for your answer, that makes total sense. Although someone commented on my original question saying it fails under a certain condition (the one with the 'D'). So does or doesn't that need to hold?? – kam Jan 08 '21 at 19:22
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    This isn't correct. The answer of @hamam_Abdallah provides a counterexample. The problem is that $y$ can be arbitrarily close to $\beta$ without $f(y)$ approaching $\alpha$. You need some kind of continuity condition to complete the proof. – Robert Shore Jan 08 '21 at 19:42
  • @RobertShore which step, explicitly, does not work? Note that his counterexample does not satisfy the condition I gave at the beginning. – Kenta S Jan 08 '21 at 19:43
  • @KentaS Ah, I see. Make some trivial edit and I'll remove the downvote. – Robert Shore Jan 08 '21 at 19:49
  • @KentaS Votes are locked after $5$ minutes, but are unlocked if the answer is edited. – Robert Shore Jan 08 '21 at 19:54
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    @RobertShore I edited it. I didn't know that, thanks! – Kenta S Jan 08 '21 at 19:55
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This is true, and the $\varepsilon-\delta$-approach isn't that tedious, actually: For every $\varepsilon>0$ there is a $\lambda>0$ such that $\vert f(x)-L\vert<\varepsilon$ for all $x$ with $\vert x-\alpha\vert<\lambda$. Also, for each $\lambda>0$, there is a $\delta>0$ such that $\vert g(y)-\alpha\vert<\lambda$ for all $y$ with $\vert y-\beta\vert<\delta$. Put both facts together to get that for all $\varepsilon>0$ there is a $\delta>0$ such that $\vert f(g(y))-L\vert<\varepsilon$ for all $y$ with $\vert y-\beta\vert<\delta$.

Edit: To be clear, this assumes the following definition of the limit of a function. Let $X\subseteq\mathbb R$, $f:X\to \mathbb R$, $x_0\in X$. Then $\lim_{x\to x_0}f(x)=y$ if for all $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$ with $\vert x-x_0\vert<\delta$ we have $\vert f(x)-y\vert<\varepsilon$. In a more topological language, for every open neighborhood $V$ of $y$ in $X$ there is an open neighborhood $U$ of $x_0$ such that $f(U)\subseteq V$. This is an contrast to a slightly different definition where the condition need only hold for all $x\in X\backslash\{x_0\}$, or in a topological language, for all punctured neighborhoods of $x_0$. The two are not equivalent, and my argument only holds with the former. To work with the latter, $g$ must also be continuous.

Vercassivelaunos
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    This isn't correct. You can't put the facts together without some continuity condition on $g$. See the counterexample of @hamam-Abdallah. – Robert Shore Jan 08 '21 at 19:43
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    @RobertShore: Depends on your definition of limit. If it is strictly "for all $y$ with $\vert y-\beta\vert<\delta$", then $g$ is automatically continuous at $\beta$, since then the condition must also hold for $y=\beta$, so $g(\beta)=\lim_{x\to\beta}g(\beta)$. If your definition of limit is "for all $y\neq\beta$ with $\vert y-\beta\vert<\delta$", then you do need to enforce continuity. One definition has more of a topological background, the other is more grounded in calculus. – Vercassivelaunos Jan 08 '21 at 20:09