how to find epsilon-delta definition of a limit

Question

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?

Answer 1

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 $|\sin{x}|\leq 1$.)

Answer 2

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. $\square_2$