Suppose f is differentiable on the interval [a, infinity) and the limit of f as x approaches a from the right equals the limit of f as x approaches infinity, both equal to a finite number A. Prove that there exists xi in (a, infinity) with f'(xi) = 0.

Step-by-step solution

1. 1

   Recognise this as Rolle's theorem stretched over an infinite interval. Rolle needs equal values at two endpoints; here you only have equal limits, and one end is at infinity, so the theorem does not apply directly. The workaround is Fermat's theorem instead: at an interior point where a differentiable function attains a local extremum, the derivative is zero. The whole proof therefore reduces to trapping an extremum strictly inside $(a,+\infty)$.

2. 2

   Dispose of the constant case. If $f(x)\equiv A$ on $(a,+\infty)$, then $f'(x)=0$ for every $x$, and any $\xi$ works. So assume from now on that there is a point $x_{0}\in(a,+\infty)$ with $f(x_{0})\neq A$. Replacing $f$ by $-f$ if necessary, we may assume

   $f(x_{0})>A$

   since $-f$ has the same limits (negated) and the same zero-derivative points.

3. 3

   Use the two limits to confine large values to a bounded window. Put

   $\varepsilon=\frac{f(x_{0})-A}{2}>0$

   From $\lim_{x\to a^{+}}f(x)=A$ there is $u\in(a,x_{0})$ such that $|f(x)-A|<\varepsilon$, hence $f(x)<A+\varepsilon$, for all $x\in(a,u]$. From $\lim_{x\to+\infty}f(x)=A$ there is $v>x_{0}$ such that $f(x)<A+\varepsilon$ for all $x\geq v$. So outside $[u,v]$ the function stays below $A+\varepsilon$, while inside it reaches

   $f(x_{0})=A+2\varepsilon>A+\varepsilon$

4. 4

   Attain the maximum on the closed interval and show it is interior. $f$ is differentiable, hence continuous, on the compact interval $[u,v]$, so by the extreme value theorem it attains a maximum there at some $\xi\in[u,v]$. That maximum is at least $f(x_{0})=A+2\varepsilon$. But $f(u)<A+\varepsilon$ and $f(v)<A+\varepsilon$, so $\xi$ is neither endpoint:

   $\xi\in(u,v)\subset(a,+\infty)$

5. 5

   Upgrade the maximum on $[u,v]$ to a global maximum, then apply Fermat. For $x\in(a,u)$ and for $x>v$ we have $f(x)<A+\varepsilon<f(\xi)$, so $f(\xi)$ is in fact the largest value of $f$ on all of $(a,+\infty)$; in particular it is a local maximum at an interior point. Since $f$ is differentiable at $\xi$, Fermat's theorem gives

   $f'(\xi)=0$

   which completes the proof.

6. 6

   Note where each hypothesis was needed. Finiteness of $A$ was essential for choosing $\varepsilon$; equality of the two limits is what makes the function return to the same level at both ends so a bump in the middle cannot be a monotone trend; and differentiability on all of $(a,+\infty)$ is what lets Fermat's theorem fire at $\xi$. Drop any one of them and the conclusion can fail — for instance $f(x)=1/x$ on $[1,\infty)$ has different end limits and no critical point.