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For $f(x)=\sin(x)\sin(\frac{1}{x})$ as $x\to 0 $.

limsup should be $0$ and liminf should be $1$. Is my answer correct?

Departed
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2 Answers2

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No.

$|f(x)| \le | \sin x|$ for all $x$. Hence $f(x) \to 0$ for $x \to 0$.

Fred
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You have here an example of the

Claim: if $\;\lim\limits_{x\to x_0}g(x)=0\;$ and $\;h(x)\;$ is a function defined in some neighborhood of $\;x_0\;$ and bounded there , then $\;\lim\limits_{x\to x_0}g(x)h(x)=0\;$

Thus, the limit of your function is zero, and then the lim sup and the lim inf. equal zero, too.

DonAntonio
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  • Hints to prove this claim? – Praveen Dec 21 '16 at 16:07
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    @supremum Oh, it's very simple: suppose $;|h(x)|<M;$ in the said nieghborhood, and take any $;\epsilon>0;$ . Then, there exists $;\delta>0;$ s.t. $$|x|<\delta\implies |g(x)|<\frac\epsilon M$$ so for this same $;|x|<\delta;$ we get$$|g(x)h(x)|=|g(x)||h(x)|<\frac\epsilon M\cdot M$$ and we're done... – DonAntonio Dec 21 '16 at 16:41
  • You mean $|x-x_0|<\delta$? – Praveen Dec 21 '16 at 16:51
  • @supremum Yes, of course. I just got distracted by your limit, but it is similar – DonAntonio Dec 21 '16 at 17:20