Prove that limit doesn't exists at all I have a function $f(x)=\frac{1}{x}+\sin(\frac{1}{x})$ which I have proved that $\lim_{x \to0^+}f(x)=\infty$, and I need to prove that $\lim_{x \to0^+}f'(x) \ne -\infty$.
I have got that $f'(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$
I am trying to prove that:
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$$
Does not equal to $-\infty$ (or to prove that it doesn't exists at all).
I have tried to assume by contradiction that
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\infty$$
So there exists $\delta_{1}>0$ such that for every $0<x<\delta_{1}$ we say that $$f(x)<\square $$
We choose $\delta=\delta_{1}$, and let $0<x<\delta$ so:
$$f(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\frac{1}{x^2}(1+\cos(\frac{1}{x}))\le -\frac{2}{x^2}<\frac{2}{\delta^2}=k$$
I do not know what to choose instead of $\square$, and if what I did is correct.
Appreciate your help!
I have also tried to think about the continuity of $f$, and saying that it is not bounded so it can't be continuous, so the limit doesn't exists.
Thanks a lot!
 A: Because there exists a subsequence converging to $0$:
if you pick $\{x_k\}$ to be such that $x_k=\frac{1}{\pi + 2\pi k}$, then
$$f^\prime (x_k) = -(\pi + 2\pi k)^2 \Big(1+cos(\pi + 2\pi k)\Big) = 0,$$
since
$$cos(\pi + 2\pi k)=-1 \mbox{, } \forall k;$$
Also the sequence of $x_k$ converges to $0$ as you can observe letting $k\to +\infty$.
A: With $z:=\dfrac1x$, rewrite as
$$\lim_{z\to\infty}-z^2(1+\cos z).$$
Then it is clear that the function alternates between $0$ and $-2z^2$.
A: Scrap the whole proof by contradiction thing, it's not necessary.
Hint: Find the values of $x$ such that $\frac{1}{x} = 2k\pi$, and $\frac{1}{x} = (2k+1)\pi,$ where $k \in \mathbb{Z}$.
Edit: In fact, as Calvin Khor pointed out in the comments to this answer, you only need one of the above two sequences to show that $\lim_{x \to0^+}f'(x) \ne -\infty$, which was the original question.
Further edit: I think OP doesn't properly understand what is meant by $lim_{x \to c^+}g(x) = -\infty$. I also needed a sanity check, which is why I posted this question.
If $g(x)$ is a real function, then a definition of $\lim_{x \to c^+}g(x) = -\infty$ is:

for each $\gamma \in \mathbb{R}, \exists \delta > 0 $ such that $x \in (c,c+\delta) \implies g(x) < \gamma$.

So an equivalent definition of $\ \neg\left(\lim_{x \to c^+}g(x) = -\infty\ \right) $ is:

there exists $\beta\in\mathbb{R}$ such that for every $\delta>0$, there exists $x'\in(c,c+\delta)$ with $g(x')\ge \beta$

If no such $\beta$ exists, then $\ \lim_{x \to c^+}g(x) = -\infty\ .$
Note that $g$ need not be continuous at $c$ in either of my definitions.
A: Part 1:
You can try making it a bit "easier" by substitution.
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{cos(\frac{1}{x})}{x^2}\underset{t=\frac{1}{x}}{\Rightarrow}$$
$$\lim_{t\to\infty}-t^{2}\left(1+\cos t\right)$$
Now it should be easier to negate by definition, as you can easily see that no matter how big $t$ is, and how small $f'$ is for those big $t$'s, you can choose $t<t_1$ such that $f'(t_1)=0$
If you'll still be struggling, make sure to leave a comment and I will help you further.
Part 2:
Assuming, in contradiction that $\lim_{t\to\infty}-t^{2}\left(1+\cos t\right)=-\infty$
Then there exists $M_{1}<0$ such that for every $t<M_{1}$ we have:$$f\left(t\right)<-1$$
Choosing $M_{1}>t_{1}=\lfloor M_{1}\rfloor\cdot2\pi-\pi$ , we get: $$f\left(t\right)=0>-1$$ Contradiction!
A: Assume by contradiction that for every $K<0$ there exists a $\delta>0$ such that for every $0<x<\delta$, we get $f'(x)<K$.
We choose $n \in \mathbb{N}$ such that $n>K$, and we choose $x_{n}= \frac{1}{\pi +2 \pi n}$.
Note that $0<\frac{1}{\pi +2 \pi n}<1$. (So in this case $\delta=1$)
And: $f'(x)=-(\pi +2\pi n)(\cos(\pi+2\pi n)+1)=0>K$
So $f'(x)>K$, in contradiction!
Is that correct?
Thanks!
