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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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 $a$.
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 $a$.
SAMPLE SOLUTION
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Note that this solution breaks the problem into two cases to show when we can say that
<Sign in to see all the formulas>
This is not necessary to do the proof. The more direct and simpler proof can be found here.
You may want to recall the definitions of function composition (<Sign in to see all the formulas>) and continuous functions.
To do this proof we will examine two cases. One in which an additional assumption is true and one in which it is not true. So for the first case, I am going to assume that there exists some <Sign in to see all the formulas> such that <Sign in to see all the formulas> for all <Sign in to see all the formulas>.
Case 1:
Let's be clear about our assumptions and what they mean in terms of inequalities:
Now since we choose <Sign in to see all the formulas> and <Sign in to see all the formulas>, let's set <Sign in to see all the formulas> and <Sign in to see all the formulas>. Let's choose a value for <Sign in to see all the formulas> and assume that we are at an $x$ such that <Sign in to see all the formulas> then <Sign in to see all the formulas>. Let's further restrict $x$ such that <Sign in to see all the formulas>. So now if <Sign in to see all the formulas> then <Sign in to see all the formulas> by the third assumption.
Let's assert that the function limit of <Sign in to see all the formulas> is
<Sign in to see all the formulas>
and see if it is true. For the limit to be true we need to show that for every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that if <Sign in to see all the formulas> then <Sign in to see all the formulas>. Let's use the same (arbitrary) <Sign in to see all the formulas> and <Sign in to see all the formulas> as above. Now assume that we are at $x$ such that <Sign in to see all the formulas>. As we showed above, <Sign in to see all the formulas> and from the second assumption above <Sign in to see all the formulas>. Therefore,
<Sign in to see all the formulas>
It is important to note that finding the <Sign in to see all the formulas> as we did allows us to equate
<Sign in to see all the formulas>
when <Sign in to see all the formulas>. This is because we restricted <Sign in to see all the formulas> so that $x$ would be in an interval where <Sign in to see all the formulas>. To see this note that when <Sign in to see all the formulas>, <Sign in to see all the formulas> and therefore <Sign in to see all the formulas>. By definition this is the limit
<Sign in to see all the formulas>
Case 2:
Now let's take the second case and assume that our assumption is not true. This means that for every <Sign in to see all the formulas> there exists an $x$ in <Sign in to see all the formulas> such that <Sign in to see all the formulas>. For the times when <Sign in to see all the formulas>, <Sign in to see all the formulas> is zero and therefore less than an arbitrary <Sign in to see all the formulas> we may choose. For those values of $x$ in <Sign in to see all the formulas> where <Sign in to see all the formulas> then the argument above holds. The difference between the two cases is that in the second we can't say that
<Sign in to see all the formulas>
because there are times when <Sign in to see all the formulas>. However, in both cases we can say that
<Sign in to see all the formulas>
Please note that we do not have to use these two cases to prove that
<Sign in to see all the formulas>
However, if we want to say that
<Sign in to see all the formulas>
then we can only do so when <Sign in to see all the formulas>. In other words, just because we know that $g(x)$ goes to $g(a)$ when $x$ goes to $a$, we cannot always say that
