# 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 $$y$$.

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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