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February 18th, 2015, 09:20 AM   #1
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Propositions concerning a differentiable function.

I am trying to solve the following problems:

Suppose that f: (0, ∞) → R is differentiable. Determine whether each of the following statements is true or false:

(a) If limx→∞ f(x) exists and is finite, and limx→∞ f'(x) = b, then b = 0.

(b) If limx→∞ f(x) exists and is finite, then limx→∞ f'(x) = 0.

(c) If limx→∞ f'(x) = 0, then limx→∞ f(x) exists.

(d) If limx→∞ f'(x) = 0, then limx→∞ f(x)/x = 0.

I have drawn pictures for (a), and I think that it is true. However, I'm not sure how to prove this. I've managed to find a counterexample for (c).
I'm not sure how to approach (b) and (d), although I have noticed that (a) and (b) are related.

Could I have suggestions on how to approach these questions?
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February 18th, 2015, 10:27 AM   #2
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a) Suppose that $b \ne 0$. Consider the $\delta-\epsilon$ definition of the limit of $f$ and come to a contradiction.

b) What if the limit doesn't exist?
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February 18th, 2015, 04:54 PM   #3
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Recall that $$\lim_{x \to \infty} f'(x) = b$$means that for all $\epsilon \gt 0$ there exists a $\delta$ such that $$x \gt \delta \implies |f'(x) - b| \lt \epsilon$$
Thus we have
$$\newcommand{\dt}{\,\mathrm d t}\begin{aligned} && -\epsilon &\lt f'(t) - b \lt \epsilon &\text{for all $t \gt \delta$} \\ && \int_\delta^x -\epsilon \dt &\lt \int_\delta^x f'(t) - b \dt \lt \int_\delta^x \epsilon \dt \\ && (\delta - x)\epsilon &\lt f(x) - f(\delta) - b(x - \delta) \lt (x - \delta)\epsilon \\ &\text{gathering all the constant terms into one, we get} & (b-\epsilon) x &\lt f(x) - L \lt (b + \epsilon)x &\text{(equation 1)}\end{aligned}$$

Now we address part a)

When $b \gt 0$ we take the left hand inequality of equation 1, with $\epsilon = {b \over 2}$ (this sets the value of $L$, but it is still a constant) to get
$${b \over 2}x + L \lt f(x)$$and since the left hand side grows without bound, so does $f(x)$. This means that $f(x)$ has no limit, contradicting the assumptions of proposition a).

When $b \lt 0$ we take the right hand inequality of equation 1, with $\epsilon = -{b \over 2}$ to get $$f(x) \lt L + {b \over 2}x$$and since $b \lt 0$ the right hand side decreases without bound and therefore $f(x)$ does too contradicting the assumptions of proposition a).

Thus $$\lim_{x \to \infty} f'(x) = b \implies b = 0$$

Proposition b) omits the assumption that the limit exists. So we may search for a counter example where there is no limit.
$$f(x) = c + {1 \over x} \sin x^2 \implies \lim_{x \to \infty} f(x) = c \\ f'(x) = {1 \over x}2x \cos x^2 -{1 \over x^2}\sin x^2 = 2\cos x^2 -{1 \over x^2}\sin x^2$$
The second term of the derivative converges to zero, but the first oscillates between 0 and 1. So $f'(x)$ does not tend to a limit as $x \to \infty$.

For proposition c) $f'(x) = {1 \over x}$ is a counter-example.

For proposition d) we return to equation 1 and divide by $x$ to get
$$b - \epsilon \lt {f(x) \over x} - {L \over x} \lt b + \epsilon \\ {L \over x} -\epsilon \lt {f(x) \over x} - b \lt {L \over x} + \epsilon$$
And this is valid for all $x \gt \delta$ so $|{L \over x}| \lt |{L \over \delta}|$ and thus we have $$\left|{f(x) \over x} - b\right| \lt \left|{L \over \delta}\right| + \epsilon$$ both terms on the right can be made arbitrarily small by choosing small enough $\epsilon$ and large enough $\delta$ so we have the convergence of ${f(x) \over x}$ as required (note that $b=0$ in the particular case given).
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February 19th, 2015, 09:45 PM   #4
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There are some errors in the work above (although the ideas are correct). In particular equation (1) is incorrect.

Recall that $$\lim_{x \to \infty} f'(x) = b$$means that for all $\epsilon \gt 0$ there exists a $\delta$ such that $$x \gt \delta \implies |f'(x) - b| \lt \epsilon$$
Thus we have
$$\newcommand{\dt}{\,\mathrm d t}\begin{aligned} && -\epsilon &\lt f'(t) - b \lt \epsilon &\text{for all $t \gt \delta$} \\ && \int_\delta^x -\epsilon \dt &\lt \int_\delta^x f'(t) - b \dt \lt \int_\delta^x \epsilon \dt \\ && (\delta - x)\epsilon &\lt f(x) - f(\delta) - b(x - \delta) \lt (x - \delta)\epsilon &\text{(equation 1)}\end{aligned}$$
When $b \gt 0$ we take the left hand inequality of equation (1), with $\epsilon = {b \over 2}$ to get
$${b \over 2}x + L \lt f(x) \qquad (L = f(\delta) - \tfrac{b}{2}\delta \quad \text{a constant})$$and since the left hand side grows without bound, so does $f(x)$. This means that $f(x)$ has no limit, contradicting the assumptions of proposition a). And similarly the right hand inequality shows that $b \lt 0$ implies that $f(x)$ is unbounded below.

For proposition d) we return to equation (1), taking $b=0$, and divide by $x$ to get
$$\left({\delta \over x} - 1\right)\epsilon \lt {f(x) - f(\delta) \over x} \lt \left(1 - {\delta \over x}\right)\epsilon$$
Since the equation is valid for $x \gt \delta$ we have ${\delta \over x} \lt 1$ and so we can write $1 - {\delta \over x} > p > 0$ for large enough $x$ and we can thus write $p\epsilon = \epsilon'$ to get
$$\begin{aligned} \epsilon' \lt {f(x) - f(\delta) \over x} &\lt \epsilon' \\ \implies \left| {f(x) - f(\delta) \over x} \right| &\lt \epsilon' \\ \text{so} \quad \lim_{x\to \infty} {f(x) - f(\delta) \over x} &= 0 \\ \lim_{x\to \infty} \left({f(x) \over x} - {f(\delta) \over x}\right) &= 0 \\ \lim_{x\to \infty} {f(x) \over x} - \lim_{x\to \infty} {f(\delta) \over x} &= 0 \\ \lim_{x\to \infty} {f(x) \over x} &= \lim_{x\to \infty} {f(\delta) \over x} = 0 \\ \end{aligned}$$
as required.
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