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I was trying to make sense of the Limit of a Composite function and (through stack exchange) found that: If $\lim_{u\to b} f(u) = c$ and $\lim_{x\to a} g(x) = b$, then $\lim_{x\to a} f(g(x)) = c$, as long as f(x) is continuous at b. This, I can make sense of.

However, looking at the proof given on ProofWiki (https://proofwiki.org/wiki/Limit_of_Composite_Function), it is stated that this condition of continuity does not need to be met as long as "for some open interval $I$ containing $\xi$, it is true that $g(x)\neq \eta$ for any $x\in I$ except possibly $x=\xi$" where "$\xi$" refers to "$a$" and "$\eta$" refers to "$b$" above.

Why is this the case? I understand that this is necessary for the proof and for the result to hold, but can anyone offer an explanation as to why satisfying this condition ensures that the result holds in a more conceptual/general sense? There is likely some aspect of limits of composite functions that I'm failing to picture here. I can't pin-point it, but maybe it has to do with the limit with $g(x) \to b$ approaching from above or below?

carmichael561
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Quantum Dot
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  • I really don't see why people bring continuity into picture whenever there is a discussion of limit of composite functions (or rule of substitution in limits). I have mentioned several times on MSE but it appears that it is not enough. See this answer http://math.stackexchange.com/a/1726877/72031 – Paramanand Singh Jun 04 '16 at 00:38
  • It is better if you think informally (meaning rigorously without using formal symbols) about limits. Since $g(x) \to b$ as $x \to a$ it follows that values of $g(x)$ can be ensured to be near $b$ by keeping values of $x$ near $a$ but not equal to $a$. Again by same logic values of $f(x)$ can be ensured to be near $c$ by keeping values of $x$ near $b$ but not equal to $b$. This means that values of $f(g(x))$ can be ensured to be near $c$ if we can ensure values of $g(x)$ to be near $b$ (but not equal to $b$) and this is ensured by our hypotheses by keeping values of $x$ near $a$. Contd. – Paramanand Singh Jun 04 '16 at 00:49
  • You see that we don't need continuity of $f$ anywhere in the above logic. What we really need to ensure is that $g(x) \neq b$ for all $x$ sufficiently near $a$ (but not equal to $a$). – Paramanand Singh Jun 04 '16 at 00:51
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    Ah I see. Thank you for the conceptualization. It is like a consequence of a standard limit x-->a of f(x) = L in that not only do we not care about the behavior at x = a, but we must disregard it entirely because the function may behave differently at that point (unless continuity at said point is guaranteed). When we deal with composite functions, this becomes apparent with the outer function and presents issues if we do not ensure that the inner function does not equal the limit in some neighborhood containing a. At least that is how I rationalized your response. Thank you. – Quantum Dot Jun 05 '16 at 22:52
  • You have understood my comment perfectly!! +1 for your understanding. – Paramanand Singh Jun 06 '16 at 04:38
  • @ParamanandSingh While it is true that you can formulate the theorem without continuity of $f$, it will turn out to be always the case that you cannot find a counterexample where $f$ is continuous at $b$. – Maxis Jaisi Sep 07 '19 at 13:13
  • @MaxisJaisi: the theorem works for both continuous and discontinuous functions. If the function $f$ is discontinuous, one needs to assume that $g(x) \neq b$. If $f$ is continuous then this assumption is not needed. So there is no reason to find counter-example for the case when $f$ is continuous. – Paramanand Singh Sep 08 '19 at 04:19

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