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$\lim \limits_{x \to 0}x\sin\left(\frac{1}{x}\right)$

I need to find and prove this limit. I can easily plug it into wolfram alpha, but I want to make sure I learn something. It's been 3 years since my Calculus 2 course and I just cannot remember how to tackle this one.

Hints?

nullByteMe
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4 Answers4

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Hint $|x \sin(1/x)|$ $\leq |x|| \sin(1/x)|$

SquirtleSquad
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As $x$ approaches zero, $1/x$ approaches infinity.

The point here is that $\sin(1/x)$ remains bounded all the time, so in the product $x\sin(1/x)$ the only thing that really matters is the $x$ part.

Daniel
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  • Since it's bounded by $0$ I can merely substitute in $0$, which gives me a limit of $1$? – nullByteMe Jul 02 '16 at 00:22
  • $\sin(1/x)$ is not bounded by zero, this function is bounded by $-1$ and $1$. At the end, you need to take absolute values (as Merlin suggests in his answer) and then the $\mid\sin(1/x)\mid$ part will become less than or equal to $1$, so in the inequality the only remaining thing will be the $\mid x \mid$, and we know its limit... – Daniel Jul 02 '16 at 00:25
  • Oh so this is much easier than I was making it out to be since the llimit of $|x|=0$ making the entire thing $0$ – nullByteMe Jul 02 '16 at 00:32
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Apply "squeeze theorem"

$$0\le \left| x\sin { \left( \frac { 1 }{ x } \right) } \right| \le \left| x \right| $$

haqnatural
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One way you can do this limit is through a substitution. Let $x=\frac{1}{t}$. The limit becomes very easy then: A bounded sine term in the numerator and a linear $t$ in the denominator and the latter goes to infinity, thus...

imranfat
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  • I can just pull out the $\frac{1}{x}$ and make $\frac{1}{t}$ and take $t$ to infinity and sub that back in? – nullByteMe Jul 02 '16 at 00:24
  • The $\frac{1}{x}$ is replaced by $t$ and so that becomes $sint$. The $x$ that sits upfront is multiplied wit the sine term, but the $x$ is now $\frac{1}{t}$. So you get $\frac{sint}{t}$ where $t$ goes to (negative) infinity. The answer thus becomes zero – imranfat Jul 02 '16 at 00:26
  • That substitution is not possible, since as $x\to 0$, $t$ either has to go to $\infty$ or $-\infty$. – Workaholic Jul 02 '16 at 00:30
  • Of course that substitution is possible. You just got to consider two cases and both cases are well defined and have the same answer. These kind of situations are not uncommon in limits like this. – imranfat Jul 02 '16 at 00:31
  • @imranfat But you should have been explicit on that, using $x=\tfrac 1t$ alone with the limit $\lim\limits_{x\to0}x\sin\left(\tfrac1x\right)$ wont make sense unless we consider two separate cases, when $x\to0^+$ and when $x\to0^-$. – Workaholic Jul 02 '16 at 00:33
  • Yes, but I kind of left it to the OP to realize. He has taken Calculus 2 before, so he is not a newbee...In Calc2, this is not a big thing. – imranfat Jul 02 '16 at 00:35
  • For my proof I came up with the following: $\left|\sin\left(\frac{1}{x}\right)\right| \lt \delta = \epsilon$. Obviously I left out a few steps, but does this seem correct? – nullByteMe Jul 02 '16 at 01:05