Prob. 16 (c), Sec. 6.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: If $f^\prime(x)\to b$ as $x \to \infty$, then $f(x)/x \to b$. How?

Question

Here is Prob. 16, Sec. 6.1, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

Let $f \colon [0, \infty) \to \mathbb{R}$ be differentiable on $(0, \infty)$ and assume that $f^\prime(x) \to b$ as $x \to \infty$.

(a) Show that for any $h > 0$, we have $\lim_{x \to \infty} \big( f(x+h) - f(x) \big)/h = b$.

(b) Show that if $f(x) \to a $ as $x \to \infty$, then $b = 0$.

(c) Show that $\lim_{x \to \infty} \big( f(x)/x \big) = b$.

Here is Definition 4.3.10 in Bartle & Sherbert, 4th edition:

Let $A \subseteq \mathbb{R}$ and let $f \colon A \to \mathbb{R}$. Suppose that $(a, \infty) \subseteq A$ for some $a \in \mathbb{R}$. We say that $L \in \mathbb{R}$ is a limit of $f$ as $x \to \infty$, and write $$ \lim_{x \to \infty} f = L \qquad \mbox{ or } \qquad \lim_{x \to \infty} f(x) = L, $$ if given any $\varepsilon > 0$ there exists $K = K(\varepsilon) > a$ such that for any $x > K$, then $\lvert f(x) - L \rvert < \varepsilon$.

I think I'm clear on Parts (a) and (b).

How to do Part (c)?

Here is my attempt at Part (c):

As $f^\prime$ is defined for all $x$ in $(0, \infty)$ and as $f^\prime(x) \to b$ as $x \to \infty$, so given any $\varepsilon > 0$ we can find a $K \colon= K(\varepsilon/3) > 0$ such that for any $x > K$, we have $$ \left\lvert f^\prime(x) - b \right\rvert < \frac{\varepsilon}{3}. \tag{0}$$

In particular, for $\varepsilon = 1$, there exists a real number $K_0 \colon= K(1) > 0$ such that for any $x > K_0$, we have $$ \left\lvert f^\prime(x) - b \right\rvert < 1.$$ So for any $x > K_0$, we also have $$ \left\lvert f^\prime(x) \right\rvert \leq \left\lvert f^\prime(x) - b \right\rvert + \lvert b \rvert < 1 + \lvert b \rvert, $$ and thus $$ \left\lvert f^\prime(x) \right\rvert < 1 + \lvert b \rvert \tag{1} $$ for any real number $x > K_0$.

Now let $M$ be any real number such that $$ M > \max \left\{ \ K, K_0 \ \right\}. \tag{2}$$

Let us take $x$ to be any real number such that $$x > \max\left\{ \ M, \frac{3 \big( \lvert f(M) \rvert + 1 \big) }{\varepsilon }, \frac{3 M \big( \lvert b \rvert + 1 \big)}{ \varepsilon } \ \right\}. \tag{3}. $$

Then as $f$ is continuous on the closed interval $[ M, x]$ and differentiable on the open interval $( M, x)$, so there exists a real number $c_x \in (M, x)$ such that $$ f(x) - f(M) = (x-M) f^\prime \left( c_x \right), $$ and then $$ f(x) = f(M) + (x-M) f^\prime \left( c_x \right), $$ which implies that $$ \begin{align} & \ \ \ \left\lvert \frac{f(x)}{x} - b \right\rvert \\ &= \left\lvert \frac{ f(M) + (x-M) f^\prime \left( c_x \right) }{x} - b \right\rvert \\ &= \left\lvert f^\prime \left( c_x \right) - b + \frac{f(M)}{x} - \frac{M f^\prime\left(c_x\right)}{x} \right\rvert \\ &\leq \left\lvert f^\prime \left( c_x \right) - b \right\rvert + \frac{ \left\lvert f(M) \right\rvert }{x} + \frac{ M \left\lvert f^\prime\left(c_x\right) \right\rvert }{x} \\ & \ \ \ \mbox{ [ Note that our choice of $K$, $K_0$, and $M$ in (0), (1), and (2) above, respectively, implies that $x > 0$ also. ] } \\ &< \frac{\varepsilon}{3} + \frac{ \left\lvert f(M) \right\rvert }{x} + \frac{M}{x} \left( \lvert b \rvert + 1 \right) \\ & \ \ \ \mbox{ [ using (2), and then (0) and (1) above ] } \\ & < \frac{\varepsilon}{3} + \frac{\lvert f(M)\rvert \varepsilon}{3 \big( \lvert f(M) \rvert + 1 \big) } + \frac{\varepsilon}{3 \left( \lvert b \rvert + 1 \right) } \left( \lvert b \rvert + 1 \right) \\ & \qquad \qquad \mbox{ [ using (3) above ] }\\ &\leq \frac{\varepsilon}{3} + \frac{ \varepsilon }{ 3 } + \frac{\varepsilon }{3} \\ &= \varepsilon. \end{align} $$

Let us put $$ M^* \colon= \max\left\{ \ M, \frac{3 \big( \lvert f(M) \rvert + 1 \big) }{\varepsilon }, \frac{3 M \big( \lvert b \rvert + 1 \big)}{ \varepsilon } \ \right\}. \tag{4}. $$

Thus we have shown that, for any given real number $\varepsilon > 0$, there exists a real number $M^* > 0$ such that $$ \left\lvert \frac{f(x)}{x} - b \right\rvert < \varepsilon $$ for every real number $x > M^*$. Therefore $$ \lim_{x \to \infty} \frac{f(x)}{x} = b, $$ as required.

Is this proof correct? If so, is it rigorous and clear enough too? Or, are there any problems in it?

Answer 1

Follow the following steps for a neater proof:

1. If $a_n=f(n+1)-f(n)$ then $a_n \to b$ by MVT.

2. It follows that $\frac {a_1+a_2+\cdots+a_n} n \to b$. Conclude that $\frac {f(n)} n \to b$ as $ n \to \infty$.

3. If $n \leq x \leq n+1$ then $\frac {f(x)} x= \frac {f(x)-f(n)} x+\frac {f(n)} n \frac n x$. Use MVT again to conclude that $|f(x)-f(n)| <|b|+1$ as long as $n$ is large enough (and $n \leq x \leq n+1$). This shows that $\frac {f(x)-f(n)} x \to 0$. Also, $\frac {f(n)} n \frac n x \to b$. Hence $\frac {f(x)} x \to b$.