Proof clarification for Question 68 of Spivak's 'Calculus' - exploring the function $f(x)=\alpha x + x^2 \sin(1/x)$ for $\alpha =1$ Question 68 from Chapter 11 of Spivak's Calculus is about interrogating the behavior of the function $f$ defined by: $f(x)=\alpha x + x^2\sin(1/x)$ for $x \neq 0$ and $f(0)=0$, for $\alpha$ values $\geq 1$.
I am having some difficulty following Spivak's logic in the solution manual.
To provide some relevant background:

*

*$f'(x)=\alpha +2x\sin(1/x)-\cos(1/x)$ for $x \neq 0$


*Spivak creates the function $g$ defined as $g(y)=2(\sin y)/y-\cos(y)$...where $h(x)=y=\frac{1}{x}$


*For any $y_0$ where $g'(y_0)=0$, we have that $$|g(y_0)|=\frac{2+y_0^2}{\sqrt{4+y_0^4}}\gt 1$$
Here is the question of interest:

Show that if $\alpha=1$, then $f$ is not [strictly] increasing in any interval around $0$.

Here is Spivak's solution manual's entry:

We have $f'(x)=1+g(1/x)$. Now we clearly have $g(y) \lt 0$ for arbitrarily large $y$ (since $g(y)$ is practically $-\cos y$ for large $y$), so for arbitrarily large $y$ we have \begin{align}g(y) \lt - \frac{2+y^2}{\sqrt{4+y^4}}\lt -1 \end{align}Thus $f'(x) \lt 0$ for arbitrarily small $x$, while we also have $f'(x) \gt 0$ for arbitrarily small $x$.


Starting with the problem statement, it seems to me that our goal should be to show that for any arbitrary $\delta \gt 0$, there is an $x \in (0,\delta)$ such that $f'(x) \lt 0$. If we do that, then we can be sure that $f$ is not strictly increasing in any neighborhood around $0$.
Given this, I really do not understand what Spivak's argument is saying. Framing this from the perspective of $y$, it seems to me the Spivak's argument is saying that for any $N \gt N_0$ (where $N_0$ denotes the 'sufficiently far away' condition), we can always find a $y \gt N$ satisfying the statement:
$$g(y) \lt - \frac{2+y^2}{\sqrt{4+y^4}}\lt -1 $$
It is not clear to me why we are in a position to assert this. Spivak is presumably making use of the statement: For any $y_0$ where $g'(y_0)=0$, we have that $$|g(y_0)|=\frac{2+y_0^2}{\sqrt{4+y_0^4}}\gt 1$$
but I cannot see how. Why should it be the case that for any sufficiently far away $N$, I can find an instance of a $y_0 \gt N$ satisfying $g'(y_0)=0$? Further, even if this is true, we only know that $|g(y_0)| \gt 1$. Why could it not be the case that for all such instances of $y_0$, we have that $g(y_0) \gt 1$?
I suspect it becomes relevant to show that these $y_0$'s are local minimum/maximum, rather than merely inflection points...but I am not certain at which point this was done by Spivak.
Thanks ~
 A: NOT AN ANSWER
I read your approach and it seems to me that it is correct, though I myself don't know if that is what the author had in mind.
I would rather follow this path. When $\alpha =1$, we have, for $x \neq 0$,
$$f'(x) = 1+2x \sin\left(\frac1{x}\right)-\cos\left(\frac1{x}\right),$$
and
$$f''(x) = 2\sin\left(\frac1{x}\right) - \frac2{x}\cos\left(\frac1{x}\right) -\frac1{x^2}\sin\left(\frac1{x}\right).$$
To show, for example, that $f(x)$ is not increasing in any right neighborhood of $0$, notice that for any $k=1,2,\dots$
$$f'\left(\frac1{2k\pi}\right)=0,$$
and
$$f''\left(\frac1{2k\pi}\right)=-4k\pi.$$
so that $\frac1{2\pi k}$ is a local maximum. This answers part (c) of the question.
When $\alpha >1$ note that, for $0< x < \frac{\alpha-1}2$,
$$f'(x) = \alpha + 2x \sin\left(\frac1{x}\right) -\cos \left(\frac1{x}\right) \geq \alpha -2x -1>0$$
which answers to question (d), i.e. for $\alpha >1$ there is a neighborhood of $0$ in which the function is strictly increasing.
A: I am not entirely certain if this is what Spivak had in mind with his proof, but the following extra information seems to validate Spivak's general approach.
We need to answer the following question:

Why could it not be the case that for all such instances of $y_0$, we have that $g(y_0) \gt 1$?


We will show that for any $N$, we can find a $y_m$ satisfying $y_m \gt N$, and $g(y_m)\lt -1$
From the prompt, we know that for any $y_0$ where $g'(y_0)=0$, we have $$|g(y_0)|=\frac{2+y_0^2}{\sqrt{4+y_0^4}}\gt 1 \quad (\dagger_1)$$
Consider Spivak's function $g(y)=2(\sin y)/y-\cos(y)$. Note that $g$ is differentiable on all of $\mathbb R \ (*)$. Therefore, on any closed interval $[a,b]$, we have a minimum $y_m$.
Consider an arbitrary $N$. Next, we will show that there is a $y^* \gt N$ such that $g(y^*) = -1$. To see this, consider when $y^*=2\pi n$ for some $n \in \mathbb N$ where $n \gt \frac{N}{2\pi}$.
Now, consider the interval $[N+\frac{y^*}{2},y^*+1]$. Suppose that $y^*$ is the minimum point for this interval; importantly, note that $y^*$ is clearly not an end point. By $(*)$ this means that $g'(y^*)=0$. However, by $(\dagger_1)$, we should have that $g(y^*) \gt 1$ or $g(y^*) \lt -1$. But we showed that $g(y^*)=-1$. Clearly, then, $y^*$ is not the minimum point of the interval $[N+\frac{y^*}{2},y^*+1]$. But if $y^*$ is the not the minimum point, then there must be a $y_m \in [N+\frac{y^*}{2},y^*+1]$ such that $g(y_m) \lt g(y^*)$ . This would imply that $y_m \gt N$ and $g(y_m) \lt -1$. This approach works for any $N$ so we can generalize $(\dagger_2)$.

Next, note the relationship between $g$ and the function $h(x)=2x\sin(\frac{1}{x})-\cos(\frac{1}{x})$. In particular, we have $g(\frac{1}{x})=h(x)$. Moreover, we have that $f'(x)=1+h(x)$...or, equivalently, $f'(x)=1+g(\frac{1}{x})$. Using $(\dagger_2)$, we will show that for any $\delta \gt 0$, there is an $x \in (0,\delta)$ such that $f'(x) \lt 0$.
Consider some arbitrary $\delta \gt 0$. We are interested in an $0\lt x \lt \delta$. It is useful to frame this as $\frac{1}{x} \gt \frac{1}{\delta}$. Applying $(\dagger_2)$, let $N=\frac{1}{\delta} \gt 0$. Then we know that there is a $y_m \gt \frac{1}{\delta}$ such that $g(y_m) \lt -1$. Let $y_m=\frac{1}{x}$...rearranging, we have that $x=\frac{1}{y_m}$.
Therefore, we can conclude that $f'(\frac{1}{y_m})=1+g(y_m) \lt 0$. Noting that $\frac{1}{y_m} \gt 0$ and $y_m \gt \frac{1}{\delta}$, we have that $0\lt \frac{1}{y_m} \lt \delta$. This argument works for any $\delta$. $\quad \square$
