If $f(x)$ is continuous and $m(x)$ is the minimum value of $f(x)$ in the closed interval $[x,a]$, then prove that \[ \lim_{x\to a^-}\,m(x)=f(a). \]

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Lesson Parent: 
Function Limits VI: Continuous Functions
Problem: 

If $f(x)$ is continuous and $m(x)$ is the minimum value of $f(x)$ in the closed interval $[x,a]$, then prove that
\[
\lim_{x\to a^-}\,m(x)=f(a).
\]

Answer: 

It is true that if $f(x)$ is continuous and $m(x)$ is the minimum value of $f(x)$ in the closed interval $[x,a]$, then \[ \lim_{x\to a^-}\,m(x)=f(a). \]

SAMPLE SOLUTION

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Solution: 

It is important to note that we use this proof primarily to prove this problem which is ultimately used to prove the Fundamental Theorem of Integral Calculus for continuous functions.

From our definition of left-sided limits we must 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>.

The important thing to note here is that $f(x)$ is continuous. This means that

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Now we know that because the limit exists the right-sided and left-sided limits exist. Since the left-sided limit exists we know 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>.

From the Extreme-Value Theorem <Sign in to see all the formulas> where $x_m$ is some point in the interval $[x,a]$ where $f(x)$ takes on the minimum value.

So now for our actual proof. Let's make our guess for <Sign in to see all the formulas> to be <Sign in to see all the formulas>. Let's choose an arbitrary value for <Sign in to see all the formulas> and continue to call it <Sign in to see all the formulas>. Now we have <Sign in to see all the formulas>. Next, let's assume that we are at a value of $x$ such that <Sign in to see all the formulas>. Because the left-hand limit exists we know that <Sign in to see all the formulas>.

Now let's look at what $x_m$ can be. If $x_m$ is such that <Sign in to see all the formulas> then <Sign in to see all the formulas>. If $x_m=a$ then <Sign in to see all the formulas>. Therefore, <Sign in to see all the formulas>.

Our proof is now complete.