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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!

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

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  • $\begingroup$ 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 $-\infty$ $\endgroup$ – Calvin Khor Jan 16 at 12:22
  • $\begingroup$ Yes that's true. $\endgroup$ – Adam Rubinson Jan 16 at 12:23
  • $\begingroup$ Thank you, what do you mean by finding $x$'s values? $\endgroup$ – MathLover Jan 16 at 13:33
  • $\begingroup$ What do you think about showing that $f'$ isn't bounded? By choosing $n \in \mathbb{N}$ such that $x=\frac{1}{\pi +2 \pi n}$? $\endgroup$ – MathLover Jan 16 at 13:40
  • $\begingroup$ $k$ 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. $\endgroup$ – Adam Rubinson Jan 16 at 13:41
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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 $k\to +\infty$.

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  • $\begingroup$ Beat you to it by about 20 seconds. I'm getting quicker at this. $\endgroup$ – Adam Rubinson Jan 16 at 12:18
  • $\begingroup$ Lol, I'm cool with taking my time. $\endgroup$ – Measure me Jan 16 at 12:19
  • $\begingroup$ Why can I choose this $x$? What is the benefit of showing that $f'(x_{k})=0$? $\endgroup$ – MathLover Jan 16 at 13:32
  • $\begingroup$ 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 $0$. $\endgroup$ – Measure me Jan 16 at 13:39
  • $\begingroup$ Cool! Thank you for that! $\endgroup$ – MathLover Jan 16 at 13:43
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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 $-2z^2$.

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

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

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  • $\begingroup$ 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? $\endgroup$ – Adam Rubinson Jan 16 at 14:49
  • $\begingroup$ I meant $x>0$ not $x>c$. $\endgroup$ – Adam Rubinson Jan 16 at 15:11
  • $\begingroup$ That for every $\delta>0$ there exists $0<x<1$, in this case, $\delta=1$ and we chose $x$ to be $x_{n}$ $\endgroup$ – MathLover Jan 16 at 15:12
  • $\begingroup$ I don't understand what you mean. If you think you should edit your answer then please do so. $\endgroup$ – Adam Rubinson Jan 16 at 15:39
  • $\begingroup$ I have edited the answer :) Thanks a lot for your time my friend! $\endgroup$ – MathLover Jan 16 at 15:56

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