Prove that $$\lim_{x\to0}\sqrt{|x|}\sin\left(\frac1x+x^{10}\right)=0.$$
How do I show in a rigorous way that this limit as $x\to 0$ equals $0$ ? Any tips or suggestions would be great!
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3$\begingroup$ Tip: $\sin$ is bounded. $\endgroup$– Start wearing purpleOct 11, 2013 at 0:58
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$\begingroup$ A word of advice: You are using the terminology incorrectly: The limit does not go anywhere, it is a fixed number. The limit equals $0$. You could say that "The function approaches $0$ as $x$ approaches $0$." $\endgroup$– Andrés E. CaicedoOct 11, 2013 at 0:59
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$\begingroup$ @O.L. since sin is bounded. the limit is 0 as x approaches 0? what about square root of x? $\endgroup$– DaftyOct 11, 2013 at 1:01
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$\begingroup$ In calculus class, this is called the "Squeeze Theorem", but a calc book might not have a proof you would like. $\endgroup$– Stefan SmithOct 11, 2013 at 2:00
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2 Answers
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Since $|\sin(anything)| \le 1$, $|\sqrt{|x|}\sin(anything)| \le |\sqrt{|x|}|$, and since $\lim_{x \to 0} |\sqrt{|x|}| = 0$ (assuming you can prove this), the limit is $0$.
answered Oct 11, 2013 at 1:04
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$sin(x)\leq 1$
$\iff$ $sin(\frac{1}{x}+x^{10})\leq 1$
$\therefore \lim = \lim(\sqrt(|x|)*1)=0$
