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Let $f:\Bbb R\rightarrow \Bbb R$ be defined by $f(x)= x+ ax\sin x$.

I would like to show that if $|a| < 1$, then $\lim\limits_{x\rightarrow\pm \infty}f(x)=\pm \infty$.

Thanks for your time.

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up vote 4 down vote accepted

We start by looking at the case when $x$ is (large) positive. The idea is that if $|a|\lt 1$, then since $|\sin x|\lt 1$, the term $ax\sin x$, even if it happens to be negative, can't cancel out the large positiveness of the front term $x$. We now proceed more formally.

Note that $|\sin x|\le 1$ for all $x$, so $x|a\sin x| \le x|a|$, and therefore $x+ax\sin x\ge x-|a|x$. So our function is $\ge (1-|a|)x$ when $x$ is positive. Since $|a|\lt 1$, the number $1-|a|$ is a positive constant. But $(1-|a|)x$ can be made arbitrarily large by taking $x$ large enough.

Similarly, let $x$ be negative. Then $x+ax\sin x=x(1+a\sin x)$. But $1+a\sin x\ge 1-|a|$, and $(1-|a|)x$ can be made arbitrarily large negative by taking $x$ negative and of large enough absolute value.

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HINT: Rewrite the function as $f(x)=(1+a\sin x)x$. Now, what is $\inf_{x\in\Bbb R}(1+a\sin x)$? What does this tell you about the smallest possible value of $|f(x)|$ in terms of $x$?

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