# 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 $$0 we say that $$f(x)<\square$$

We choose $$\delta=\delta_{1}$$, and let $$0 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.

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!

• Big hint - $\cos(\pi+2k\pi) =-1$ for all $k$ – Calvin Khor Jan 16 at 12:15
• I have try thinking about if for a while, but I do not know what to do with your hint... – MathLover Jan 16 at 13:30

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.

• Your hint gives one way to show its not convergent, but you need only the second half to show that it does not converge to $-\infty$ – Calvin Khor Jan 16 at 12:22
• Yes that's true. – Adam Rubinson Jan 16 at 12:23
• Thank you, what do you mean by finding $x$'s values? – MathLover Jan 16 at 13:33
• What do you think about showing that $f'$ isn't bounded? By choosing $n \in \mathbb{N}$ such that $x=\frac{1}{\pi +2 \pi n}$? – MathLover Jan 16 at 13:40
• $k$ represents an integer (other than $0$). Do you know how to rearrange equations? And do you know what the graph of $\cos$ looks like? Then you should try to draw the graph of $\cos\left(\frac{1}{x}\right)$ using my hint. – Adam Rubinson Jan 16 at 13:41

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 $$k\to +\infty$$.

• Beat you to it by about 20 seconds. I'm getting quicker at this. – Adam Rubinson Jan 16 at 12:18
• Lol, I'm cool with taking my time. – Measure me Jan 16 at 12:19
• Why can I choose this $x$? What is the benefit of showing that $f'(x_{k})=0$? – MathLover Jan 16 at 13:32
• If you apply the definition of limit knowing this you see it doesn't work. The idea is that you always have an $x_k$ for some $k$ such that it is as close to $0^+$ as you want, so the limit can't be $-\infty$: in particular this shows that, if it exists, it is $0$. – Measure me Jan 16 at 13:39
• Cool! Thank you for that! – MathLover Jan 16 at 13:43

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 $$-2z^2$$.

• Why is it clear? – MathLover Jan 16 at 18:32
• @MathLover: do you really need a tutorial on the cosine function ? – Yves Daoust Jan 16 at 18:47
• No, I know that $cos(?)$ is between -1 and 1, but $-z^2 \to -\infty$ so I do not understand why is it clear... – MathLover Jan 16 at 19:43
• @MathLover: isn't $1+\cos z$ between $0$ and $2$ ? – Yves Daoust Jan 16 at 20:53
• @MathLover: does $0$ tend to infinity ? – Yves Daoust Jan 16 at 21:07

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 $$t such that $$f'(t_1)=0$$

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 $$t we have:$$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!

• Thank you for that, I have already tried it before actually... – MathLover Jan 16 at 14:10
• Look at the edit – GuyPago Jan 16 at 14:32
• Awesome! Thank you so much! – MathLover Jan 16 at 14:46

Assume by contradiction that for every $$K<0$$ there exists a $$\delta>0$$ such that for every $$0, we get $$f'(x).

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!

• Your first line, "Assume by contradiction that there exists $K<0$ such that $f'(x)<K$", is meaningless. Do you mean: Assume by contradiction that there exists $K<0$ such that $f'(x)<K$ for all $x$ ? Or: Assume by contradiction that there exists $K<0$ such that $f'(x)<K$ for all $x>c$? Or something else? – Adam Rubinson Jan 16 at 14:49
• I meant $x>0$ not $x>c$. – Adam Rubinson Jan 16 at 15:11
• That for every $\delta>0$ there exists $0<x<1$, in this case, $\delta=1$ and we chose $x$ to be $x_{n}$ – MathLover Jan 16 at 15:12
• I don't understand what you mean. If you think you should edit your answer then please do so. – Adam Rubinson Jan 16 at 15:39
• I have edited the answer :) Thanks a lot for your time my friend! – MathLover Jan 16 at 15:56