The behavior of $f(x)=\alpha x+x^2\operatorname{sin}1/x$ for $x\neq 0$ near $0$, where $\alpha \ge 1$. Consider $\alpha \ge 1$. 
Let $f(x)=\alpha x+x^2\operatorname{sin}1/x$ for $x\neq 0$ and let $f(0)=0.$ In order to find the sign of $f'(x)$ when $\alpha \ge 1$ it is necessary to decide if $2x\operatorname{sin}1/x-\operatorname{cos}1/x$ is $\lt -1$ for any numbers $x$ close to $0$. It is a little more convenient to consider the function $g(y)=2(\operatorname{sin}y)/y-\operatorname{cos}y$ for $y\neq 0$; we want to know if $g(y)\lt -1$ for large $y$. This question is quite delicate; the most significant part of $g(y)$ is $-\operatorname{cos}y$, which does reach the value $-1$, but this happens only when $\operatorname{sin}y=0,$ and it is not at all clear whether $g$ itself can have values $\lt -1$. The obvious approach to this problem is to find the local minimum values of $g$. Unfortunately, it is impossible to solve the equation $g'(y)=0$ explicitly, so more ingenuity is required.
(a) Show that if $g'(y)=0,$ then 
$$\operatorname{cos}y=(\operatorname{sin}y)(\frac{2-y^2}{2y}),$$
and conclude that 
$$g(y)=(\operatorname{sin}y)(\frac{2+y^2}{2y}).$$
(b) Now show that if $g'(y)=0, then 
$$\operatorname{sin}^2y=\frac{4y^2}{4+y^2},$$
and conclude that 
$$|g(y)|=\frac{2+y^2}{\sqrt{4+y^4}}.$$
(c) Using the fact that $(2+y^2)/\sqrt{4+y^4}\gt 1$, show that if $\alpha=1$, then $f$ is not increasing in any interval around $0$.
(d) Using the fact that $\lim_{y\to \infty}(2+y^2)/\sqrt{4+y^4}=1,$ show that if $\alpha \gt 1$, then $f$ is increasing in some interval around $0$.
I've worked through (a) and (b) but I have trouble solving (c) and (d). In fact, I can't even understand what the solution is saying. Below is what is written on the solution to this problem.
(c) 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 $-\operatorname{cos}y$ for large $y$), so for arbitrarily large $y$ we have 
$$g(y)\lt -\frac{2+y^2}{\sqrt{4+y^4}}\lt -1$$
by part (b). Thus $f'(x)\lt 0$ for arbitrarily small $x$, while we also have $f'(x)\gt 0$ for arbitrarily small $x$.
First of all, I have confusion over the statement $g(y)\lt 0$ for arbitrarily large $y$. Usually I would interpret this as saying $g(y)\lt 0$ for all $y\lt a$ for some $a$, but I think in this case that is not true and the author merely means that there are negative values for very large $y$. More importantly, I don't understand why the strict inequality below follows. I think what the author suggests is since for points where $g'(y)=0$, $|g(y)|=\frac{2+y^2}{\sqrt{4+y^4}}$, and at local minima we have $g'(y)=0$, there should be arbitrarily large values less than $-\frac{2+y^2}{\sqrt{4+y^4}}$. 
However, how do we know that there are even points with $g'(y)=0$?Also, even if there are such points, how do we know that they are local minimum points, or that local minimum of this function even exists anywhere? Moreover, why is the first inequality strict?
Finally, the solution to (d) is 
We have 
$$f'(x)=\alpha + g(1/x).$$
For sufficiently large $y$ we have $g(y)\gt -\alpha.$ So for sufficiently small $x$ we have $f'(x)\gt 0.$ 
My understanding of this proof is that for large $y$ g(y) is practically $-cosy$ which is bounded below by $-1$ and since $-\alpha\lt -1$, we have $g(y)\gt -\alpha.$ From here the rest follows easily, but I don't see where the fact that $\lim_{y\to \infty}(2+y^2)/\sqrt{4+y^4}=1$ is used as the the problem said.
I would greatly appreciate it if anyone could explain these points.
 A: For better context I am copy-pasting the complete answer from https://math.stackexchange.com/a/691328/72031

Clearly we have $f'(x) = 2x\sin (1/x) + 1 - \cos (1/x)$ and let's put $y = 1/x$ so that $f'(x) = (2/y)\sin y - \cos y + 1 = g(y) + 1$. Now as $x \to 0^{+}$ we have $y \to \infty$. We will show that as $x \to 0^{+}$ there are infinitely many values of $x$ for which $f'(x) < 0$. This involves that as $y \to \infty$ there are infinitely many values of $y$ for which $g(y) < -1$.
We need to analyze the maxima/minima of $g(y)$. Now $g'(y) = (2/y)\cos y - (2/y^{2})\sin y + \sin y$ and hence $g'(y) = 0$ implies that $$\cos y = (\sin y)\left(\frac{2 - y^{2}}{2y}\right)\tag{1}$$ and then at these points of extrema we have $$g(y) = \frac{2\sin y}{y} - \cos y = \frac{2\sin y}{y} - (\sin y)\left(\frac{2 - y^{2}}{2y}\right) = (\sin y)\left(\frac{2 + y^{2}}{2y}\right)\tag{2}$$ Note that from graphical considerations it is easy to show that equation $(1)$ has infinite many solutions and the values $y$ satisfying this equation increase without bound. In other words there are many large values of $y$ (exceeding any given bound) which satisfy equation $(1)$.
Next from $(1)$ we can see that $$\sin^{2}y = \frac{1}{1 + \cot^{2}y} = \dfrac{1}{1 + \left(\dfrac{2 - y^{2}}{2y}\right)^{2}} = \frac{4y^{2}}{y^{4} + 4}$$ and then from $(2)$ we get $$g(y) = \pm\frac{2 + y^{2}}{\sqrt{y^{4} + 4}}$$ where the sign $\pm$ is chosen same as the sign of $\sin y$. These values of $g(y)$ are its extrema. Note that in absolute value the expression for $g(y)$ is always greater than $1$ if $y > 0$. Hence it follows that there will be infinitely many values of $y$ corresponding to solutions of equation $(1)$ for which $\sin y$ will be negative so that $g(y) < -1$.
We have thus established that $f'(x) < 0$ for infinitely many values of $x$ as $x \to 0^{+}$. Now it is also easily seen that as $x \to 0^{+}$ the term $2x\sin (1/x) \to 0$ and $\cos (1/x)$ oscillates between $-1$ and $1$ so that the expression for $f'(x)$ oscillates between $0$ and $2$ and hence there are infinitely many values of $x$ for which $f'(x) > 0$.
Thus in any neighborhood of $0$ we have $f'(x) > 0$ for some values of $x$ and $f'(x) < 0$ for some other values of $x$. It follows that $f(x)$ is not increasing in any neighborhood of $0$.
Note: The above solution is inspired by an exercise problem (hints were provided in the problem) from "Calculus" by Michael Spivak.

Update OP's comments indicate that he has some trouble understanding the fact that there are an infinity of solutions to $(1)$ and he does not want to go for graphical considerations. Equation $(1)$ can be recast in the form $$\tan y = \frac{2y }{2 - y^{2}} \tag{3}$$ and clearly RHS tends to $0$ as $y \to \infty$. Also the RHS is negative when $y > \sqrt{2}$. Let's analyze the behavior of fraction $2y/(2 - y^{2})$ for $\dfrac{3\pi}{2} + 2n\pi < y < 2\pi + 2n\pi$ where $n$ is non-negative integer. Clearly if $$h(y) = \tan y - \dfrac{2y}{2 - y^{2}}$$ then $h(2\pi + 2n\pi) > 0$ and as $y \to (3\pi/2 + 2n\pi)^{+}$ $\tan y \to -\infty$ and so that $h(y)$ is negative as $y \to (3\pi/2 + 2n\pi)^{+}$. It follows by intermediate value theorem for continuous functions that $h(y)$ vanishes at least once in the interval $\left(\dfrac{3\pi}{2} + 2n\pi, 2\pi + 2n\pi\right)$. Since $n$ is arbitrary we get infinitely many roots of $h(y) = 0$.
Note further that for these values of $y$, $\sin y$ is negative ($y$ lies in 4th quadrant). Hence $g(y) = -(2 + y^{2})/\sqrt{y^{4} + 4} < -1$. I have no where mentioned that this $y$ is local minima. Its just local extrema. I think it is better to say that these are critical points (points where derivative vanishes) and we don't need any more information about these points in this question.
A: I thought it would be instructive to address part D, where $\alpha \gt 1$, using the information the Spivak provides the reader: $\displaystyle \lim_{y \to \infty}\frac{2+y^2}{\sqrt{4+y^4}}=1 \quad (\dagger_1)$
An important part to answering this question is demonstrating that for any $N$, we can find a $y_0 \gt N$ such that $g'(y_0)=0 \quad(\dagger_2)$. I will include the proof of this at the bottom of the post.
For an altogether easier method, consider dfnu's post on Proof clarification for Question 68 of Spivak's 'Calculus' - exploring the function $f(x)=\alpha x + x^2 \sin(1/x)$ for $\alpha =1$.

When using $(\dagger_1)$, it is important to recall the appropriate context: If $y_0$ has the property that $g'(y_0)=0$, then $|g(y_0)|=\frac{2+y^2}{\sqrt{4+y^4}}$. That is to say, an application of $(\dagger_1)$ can only be applied to critical points. When discussing critical points, I will intentionally use the symbol "$y_0$".
Now, by $(\dagger_1)$ we know that for any $\varepsilon \gt 0$, there is an $N$ such that for any $y_0 \gt N$, we have that $\left|\frac{2+y_0^2}{\sqrt{4+y_0^4}}-1\right| \lt \varepsilon$. An application of a triangle inequality variant will provide us with a more useful form: $\frac{2+y_0^2}{\sqrt{4+y_0^4}} \lt \varepsilon +1$.
Consider the particular instance where $\varepsilon = \alpha-1 \gt 0$. Then we have that for all $y_0 \gt N_{\varepsilon}: \frac{2+y_0^2}{\sqrt{4+y_0^4}} \lt \alpha $.
Equivalently, we can write this as: for any $y_0 \gt N_{\varepsilon}: |g'(y_0)|\lt \alpha$. Remember, our objective is to show that we can find an $M$ such that for any $y$ (not just critical points) greater than $M$, we have that $g(y) \gt -1$, which is derived from showing that $\alpha + g(y) \gt 0$.
Next, we will show that we can find an $M$ such that for any $y$ ( not just critical points) $\gt M$ we have $g(y) \gt -1$.
Consider two critical points $y_{0_a}$ and $y_{0_b}$ where $y_{0_a} \lt y_{0_b}$ and   $y_{0_a}, y_{0_b} \gt N_{\varepsilon}$. How do I know that two such critical points exist? Well, first, by $(\dagger_2)$ and $N=N_{\varepsilon}$, let $y_{0_a}$ be the first critical point we can find. Second, by $(\dagger_2)$ and $N=y^*\gt y_{0_a}$, let $y_{0_b}$ be the second critical point we can find.
Now, look at the closed interval $\left[y_{0_a},y_{0_b}\right]$. On this interval, $g$ is a differentiable function. Suppose there is a $y \in \left[y_{0_a},y_{0_b}\right]$ such that $|g(y)| \geq \alpha$. This would mean that there is a maximum or minimum point $y_m \in (y_{0_a},y_{0_b})$ such that $|g(y_m)| \geq \alpha$. But because $g$ is differentiable on this interval, and $y_m$ is not an end point, that means that $g'(y_m)=0$. But $y_m \gt N_{\varepsilon}$...which means $|g(y_m)| \lt \alpha$. A contradiction!
So, clearly, there are no $y \in \left[y_{0_a},y_{0_b}\right]$ such that $|g(y)| \geq \alpha$. In particular, then, there are no $y$ where $g(y) \leq -1$. Therefore, for any $y \in \left[y_{0_a},y_{0_b}\right]$, $g(y) \gt -1$.
Finally, fix $y_{0_a}=M$. It should be clear that for all $y \gt y_{0_a}$, $g(y) \gt -1 \quad \square$

$(\dagger_2)$ Proof
Consider Spivak's function $g(y)=2(\sin y)/y-\cos(y)$. Then $g'(y)=\frac{2\cos(y)}{y}-\frac{2\sin(y)}{y^2} + \sin(y)$. Some algebra produces: $$g'(y)=\frac{2\cos(y)\cdot y - \sin(y)(2-y^2)}{y^2}$$
Note that for $y\neq 0$, $g'(y)$ is a continuous function. Therefore, it is eligible for an Intermediate Value Theorem application. We will show that for any $N$, we can find a $y_1,y_2 \gt N$ where $g'(y_1) \lt 0$ and $g'(y_2) \gt 0$. Then, by the IVT, we can claim that there is a $y_0$ in the interval $(y_1,y_2)$ ( or possibly $(y_2,y_1)$ depending on whatever is larger) such that $g'(y_0)=0$. This means that for any $N$, we can find a $y_0 \gt N$ such that $g'(y_0)=0$, as desired.
Consider an arbitrary $N$. First, let us construct a $y_1 \gt N$ such that $g'(y_1) \lt 0$...i.e. find a $y_1$ such that $0\lt\frac{2\cos(y_1)\cdot y_1 - \sin(y_1)(2-y_1^2)}{y_1^2}$ Place an initial restriction that $y_1 \gt 0$. Then, some algebra shows that we need to find a $y_1$ satisfying the following condition:
$$g'(y_1)=2 \cos(y_1)-\sin(y_1)\cdot \frac{2-y_1^2}{y_1} \lt 0$$
Consider when $y_1=\pi+2\pi n$ for some yet-to-be-determined $n \in \mathbb N$. Then the above equation becomes $g'(y_1)=-2$. In order to determine the value for $n$, consider that we need $y_1 \gt N$...equivalently, we require $\pi+2 \pi n \gt N$. This implies that we need $n \gt \frac{N-\pi}{2\pi}$. The natural numbers are unbounded so this is always possible to find such an $n$.
Next, let us construct $y_2 \gt N$ such that $g'(y) \gt 0$. A similar argument will show that we can get $g'(y_2)=2$ if we let $y_2=2\pi n$ for an $n \gt \frac{N}{2 \pi}$. Applying IVT to the continuous interval $[y_1,y_2]$ (or $[y_2,y_1]$) will lead the the conclusion that there is a $y_0 \gt N: g'(y_0)=0$.This approach works for any $N$ so we can generalize. $\quad\square$
