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Showing $x^2sin frac {1}{x}$ is surjective

Mathematics Asked by Mathfun on November 26, 2020

Show that the function $f:mathbb Rrightarrow mathbb R$ defined by $f(x)=x^2sin frac {1}{x}$ if $xneq 0$ and $f(0)=0$ is surjective.

What I thought that if I could show that for every $nin mathbb N$ there is a $x in mathbb R$ such that $f(x)=n$ and there is a $y in mathbb R$ such that $f(y)=-n$ then since $f$ is continuous by IVT $f$ is surjective. But I couldn’t find such $x$ and $y$.

2 Answers

We have that

$$lim_{xto infty} x^2sin frac {1}{x}=lim_{xto infty} xfrac{sin frac {1}{x}}{frac1x} to infty$$

$$lim_{xto -infty} x^2sin frac {1}{x}=lim_{xto -infty} xfrac{sin frac {1}{x}}{frac1x} to -infty$$

and as $x to 0$

$$left|x^2sin frac {1}{x}right| to 0=f(0)$$

therefore the function is continuous and by IVT it is surjective.

Answered by user on November 26, 2020

You have that $lim_{xto +infty} f(x) =lim_{xto +infty} xcdot xsin (frac 1x) =+infty$ and in the same way $lim_{xto - infty} f(x) =-infty $, so by IVT you have thesis.

Answered by Alessandro Cigna on November 26, 2020

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