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Prove that if $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $f(g(a))$ then the function composition $(f\circ g)(x)$ is continuous at $a$ .

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Lesson Parent:

Function Limits VI: Continuous Functions

Problem:

Prove that if $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $f(g(a))$ then the function composition $(f\circ g)(x)$ is continuous at $a$ .

Answer:

It is true that if $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $f(g(a))$ then $(f\circ g)(x)$ is continuous at $a$ .

Solution:

You may want to recall the definitions of function composition (<Sign in to see all the formulas>), continuous functions, and function limits.

Let's be clear about our assumptions and what they mean in terms of inequalities:

Assumption	Meaning
<Sign in to see all the formulas>	<Sign in to see all the formulas>
<Sign in to see all the formulas>	<Sign in to see all the formulas>

Recall that both <Sign in to see all the formulas> and <Sign in to see all the formulas> are arbitrary. We are going to keep <Sign in to see all the formulas> arbitrary, but set <Sign in to see all the formulas> and at the same time set <Sign in to see all the formulas>. Now for every <Sign in to see all the formulas> we can find a <Sign in to see all the formulas> which means we can find a <Sign in to see all the formulas> such that when <Sign in to see all the formulas> then <Sign in to see all the formulas>. Now because <Sign in to see all the formulas> then <Sign in to see all the formulas>. Finally, because <Sign in to see all the formulas> then <Sign in to see all the formulas>. Again, since <Sign in to see all the formulas> then <Sign in to see all the formulas>.

Our definition of continuity of $f(x)$ at $g(a)$ assumes that we are at a value of $g(x)$ such that <Sign in to see all the formulas>. (Note that <Sign in to see all the formulas> means that <Sign in to see all the formulas>.) We have not made any constraint that <Sign in to see all the formulas>. In fact, there may be values of $x$ in <Sign in to see all the formulas> where $g(x)$ does equal $g(a)$ ! But we know that when <Sign in to see all the formulas> then <Sign in to see all the formulas> which is always less than <Sign in to see all the formulas>. In other words, our continuity of $g(x)$ at $a$ only restrains $g(x)$ to <Sign in to see all the formulas> and not <Sign in to see all the formulas>. But for those values of $x$ in <Sign in to see all the formulas> for which <Sign in to see all the formulas> then <Sign in to see all the formulas> will always be less than <Sign in to see all the formulas>.

Therefore, for every arbitrary <Sign in to see all the formulas> we can find a <Sign in to see all the formulas> such that when <Sign in to see all the formulas> then <Sign in to see all the formulas>. This is the definition of a continuous function and we have therefore proven that

<Sign in to see all the formulas>

Finally, one may be tempted to say that since $x$ goes to $a$ and $g(x)$ goes to $g(a)$ then we should be able to say that

<Sign in to see all the formulas>

We can only do this when there is some interval around

$a$ for which <Sign in to see all the formulas>. Look here for a more detailed proof.