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Use epsilon-delta definition of limit To show : $\displaystyle \lim_{x \to 0} x\sin \left(\dfrac{1}{x}\right) = 0$

How to explain this in detail with step by step explanations?

DeepSea
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rohit
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  • What have you tried? Do you know what the epsilon-delta definition of a limit actually says? – Some Math Student Nov 09 '14 at 22:50
  • no i dnt know but what to learn can u explain me step by step to solve this one? – rohit Nov 09 '14 at 22:53
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    I suggest you first look up what that definition says. Then you start thinking of, given the definition, what you have to assume and what you need to show, and start from there. – Some Math Student Nov 09 '14 at 22:54
  • yeah i studied that but didn't get a much insight into the topic. Please explain me this one – rohit Nov 09 '14 at 22:56
  • Hint: try to prove the squeeze theorem from the $\varepsilon$-$\delta$ definition of limit. Then use that in the way I stated my answer. – Matthew Levy Nov 09 '14 at 23:29

2 Answers2

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The $\varepsilon$-$\delta$-definition of a limit goes as follows:

$\lim_{x\to x_0}{f(x)}=t \iff \forall \varepsilon>0\quad \exists \delta>0$, such that $|x-x_0|<\delta \Rightarrow |f(x)-t|<\varepsilon$.

To put this differently, if you choose your $x$ closer and closer to your $x_0$, then your function value $f(x)$ will get arbitrarily close to your limit $t$.

In this case, we have $t=0$.

Now I give you, for instance, $\varepsilon=\frac{1}{2}$.

The definition tells you that, for $0$ to be the limit, you need to find a $\delta$, such that for $|x-0|=|x|<\delta$ we have $|f(x)-0|=|x\sin{\frac{1}{x}}-0|=|x\sin{\frac{1}{x}}|<\varepsilon=\frac{1}{2}$.

Can you think of a $\delta$, such that this condition is fulfilled? (A hint: note that $|\sin{x}|\leq 1$.)

Some Math Student
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Let $f(x) = x\cdot \sin\left( \frac 1 x \right)$, then

$$-1\leq \sin{\left(\frac 1 x\right)} \leq 1 \Rightarrow$$

$$-|x|\leq f(x)\leq |x|\quad \forall x\in (-1,1) \subset \Bbb R$$

So, since, $\lim \limits_{x\to 0}{|x|} = \lim \limits_{x\to 0}{-|x|} = 0$, then $\lim \limits_{x\to 0}{x\cdot \sin\left( \frac 1 x\right)} = 0$ (by the Squeeze Theorem). $\square_1$

We have that $$\forall \varepsilon > 0, \quad \exists \delta = \frac {\min\{\delta_1, \delta_2\}}{2} \text{ s.t } \left|{ x\cdot \sin \left( \frac 1 x \right)}\right| < \varepsilon \text{ when } |x|<\delta$$

which follows from our inequality above in the following way:

We know, ${ x\cdot \sin \left( \frac 1 x \right)} \leq |x| < 0+\varepsilon$ (by $x \in (-\delta , \delta)$)

and that ${ x\cdot \sin \left( \frac 1 x \right)} \geq -|x| > 0-\varepsilon$ (by $x \in (-\delta , \delta)$) $\Rightarrow$

$x\cdot \sin \left( \frac 1 x \right)\in(0- \varepsilon , 0+\varepsilon)$ when $x\in (-\delta , \delta)$.

Since this is $\forall \varepsilon > 0$ we have our proof. $\square_2$

Matthew Levy
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