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$x_{n}$ is a sequence in $(0, \infty )$ with the following attribute: $y_{n}:= x_{n} + \sqrt[5]{x_{n}} - \frac{1}{x_{n}}$ converges to an $y_{0} \in \mathbb{R}$. Proof that $x_{n}$ converges as well. Also: $f: (0, \infty ) \ni x \mapsto x + \sqrt[5]{x} - \frac{1}{x} \in \mathbb{R}$

The hints are: What's the image of $f$? Does $f$ have an inverse function which is continuous?

I have already tried to create the inverse function of $f$ and failed, did not manage to dissolve the function to $x$ because there is a root and fraction. Also because of this fraction $\frac{1}{x}$, I would say that $f$ is not continuous because we are in interval $(0, \infty )$ and division by $0$ is not possible and thus $f$ cannot have an inverse function which is continuous. With that, the first question should be done.

What's the image of $f$?

$f$: $x \rightarrow x+\sqrt[5]{x} - \frac{1}{x}$

$y$ is image of $x$ under the mapping of $f$.

So each $x$ will get (at most) one $y$ (but $x\neq 0$)

Is that correct so far? To be honest, I don't know what to do with these hints at all.

I know that $f(x)$ is not continuous but I want know for real is if $x_{n}$ converges...

berndgr
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  • is $f$ not continuous on the interval $(0,\infty)$? Well, if you can somehow show that $f^{-1}$ exists, and is continuous, then it shouldn't be hard to just use the definition of convergence of a sequence to get the desired result. – Scounged Jun 16 '16 at 16:55
  • But problem is f isn't continuous because there will be a definition gap somewhere because we cannot divide by zero, I think. If not please tell me why I'm wrong. – berndgr Jun 16 '16 at 17:03
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    remember that $0$ is not in the set $(0,\infty)$, so there will not be a definition gap on this set. – Scounged Jun 16 '16 at 17:26
  • Oh I see, thank you :) I have worked too much with [...] that's why.. – berndgr Jun 16 '16 at 18:02

1 Answers1

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Consider the function given by $f(x) = x + \sqrt[5]{x} - \frac 1x$. Then we have: $f'(x) = 1 + \frac{1}{5x^{\frac{4}{5}}} + \frac 1{x^2} > 0$. This implies that the function is invertible, since it's strictrly increasing. Here you can see the proof that this implies that the inverse function is continuous too.

Now as $f$ is continuous on $(0,\infty)$ by IVT the image of $f$ is $\mathbb{R}$. Therefore $f^{-1}$ is continuous on any point of $\mathbb{R}$. Therefore:

$$\lim_{n \to \infty} (x_n - x_0) = \lim_{n \to \infty} \big(f^{-1}(x_n + \sqrt[5]{x_n} - \frac{1}{x_n}) - f^{-1}(x_0 + \sqrt[5]{x_0} - \frac{1}{x_0})\big) = \lim_{n \to \infty} \big(f^{-1}(y_n) - f^{-1}(y_0)\big) = f^{-1}(\lim_{n \to \infty} y_n) - f^{-1}(y_0) = f^{-1}(y_0) - f^{-1}(y_0) = 0$$

Therefore $\lim_{n \to \infty} x_n = x_0$, so the sequence $(x_n)^{\infty}_{n=0}$ converges to $x_0$.

Community
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Stefan4024
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  • Thank you a lot for your answer. Besides using the first derivation, there is another way to proof that $f$ is invertible? In the end the aim of this is to proof that $f$ is monotonic, right? So I could use something else to proof that $f$ is monotonic and it would be same as yours? – berndgr Jun 16 '16 at 18:06
  • @berndgr Sort of. The idea is that a strictly increasing function has both inverse and it's continuous on image (f) – Stefan4024 Jun 16 '16 at 18:10
  • Tyvm!! I have just checked my notes and we have already luckily proofed that if a function is strictly increasing, there is an invers function :) So I can skip the proof for this. – berndgr Jun 16 '16 at 19:24