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I had few doubt about continuity of composite functions and I made a few observation about those

$1)$ if $\lim_{x \to c}f(x) = b $ and $g$ is continuous at $b$ then $\lim_{x \to c } g\circ f =g(b)$

$2)$ Given that $(g \circ f)(x) $ is continuous everywhere then $g$ is continuous at every point where $f(x)$ is defined, and coming to $f(x)$ it may be not may be continuous but it has two sided limits in its interior points.

$3)$ Given $f(x)$ is continuous at point $c$ and $g(x)$ is everywhere then concluding $g\circ f $ is continuous at every point is incorrect but at least we can say, it is continuous at $c$.

correct me if I was wrong at any point.

terran
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Luckyian
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    $g \circ f$ continuity gives nothing about $f$, in general. Consider, for example, constant $g$. Then it can has any $f$ in stomach, but composite function will be continuous. – zkutch Jan 12 '24 at 07:53
  • got it, what about other two statements, i have one more doubt, if both f(x) and g(x) are continuous at point c then we can't conclude anything about $g \circ f$ either its continuous or discontinuous. – Luckyian Jan 12 '24 at 08:00
  • Look at https://math.stackexchange.com/questions/4545377/bringing-limits-inside-functions-when-limits-go-to-infinity/4545387#4545387 – zkutch Jan 12 '24 at 08:29
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    Combining and extending your point (1) and the result zkutch linked to, it can be shown that if $\lim_{x\to c}f(x) = b$ and $\lim_{x\to b} g(x) = a$, then $\lim_{x \to c}g\circ f(x) = a$ if and only if either $g$ is continuous at $b$ ($g(b) = a$), or there is a deleted neighborhood of $c$ in which $f(x) \ne b$. – Paul Sinclair Jan 13 '24 at 13:17
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    As for (3), it can be loosened to just require $g$ to be continuous at $f(c)$, and is an immediate consequence of (1). So (2) is the only one you were mistaken on. – Paul Sinclair Jan 13 '24 at 13:19

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