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 $\delta_{1}&gt;0$ such that for every $0&lt;x&lt;\delta_{1}$ we say that $$f(x)&lt;\square $$
We choose $\delta=\delta_{1}$, and let $0&lt;x&lt;\delta$ so:
$$f(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\frac{1}{x^2}(1+\cos(\frac{1}{x}))\le -\frac{2}{x^2}&lt;\frac{2}{\delta^2}=k$$
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!
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 $k\to +\infty$.
Assume by contradiction that for every $K&lt;0$ there exists a $\delta&gt;0$ such that for every $0&lt;x&lt;\delta$, we get $f'(x)&lt;K$.
We choose $n \in \mathbb{N}$ such that $n&gt;K$, and we choose $x_{n}= \frac{1}{\pi +2 \pi n}$.
Note that $0&lt;\frac{1}{\pi +2 \pi n}&lt;1$. (So in this case $\delta=1$)
And: $f'(x)=-(\pi +2\pi n)(\cos(\pi+2\pi n)+1)=0&gt;K$
So $f'(x)&gt;K$, in contradiction!
Is that correct?
Thanks!