# Question on $\varepsilon$-$\delta$ proof for $\lim_{x \to 0} x^2\sin\left(\frac{1}{x}\right)=0$

I'm trying to understand the $$\varepsilon$$-$$\delta$$ proof for $$\lim_{x \to 0} x^2\sin\left(\frac{1}{x}\right)=0$$

We have to find $$\varepsilon>0$$ and $$\delta>0$$ such that $$\left|x^2\sin\left(\frac{1}{x}\right)\right| < \varepsilon \tag{1}\label{eq1}$$ and $$|x|<\delta \tag{2}\label{eq2}$$

My work:

We know \begin{aligned} \left|\sin\left(\frac{1}{x}\right)\right| &\leq 1 \\ \implies \left|x^2\sin\left(\frac{1}{x}\right)\right| &\leq |x^2| \\ \end{aligned} $$\therefore \text{ } \left|x^2\sin\left(\frac{1}{x}\right)\right| < \delta^2 \tag{3}\label{eq3}$$

For $$0<\delta<1$$, we have \begin{aligned} \delta^2 &< \delta \\ \implies \left|x^2\sin\left(\frac{1}{x}\right)\right| &< \delta \end{aligned} and hence we can choose any $$\delta$$, such that $$0 < \delta \leq \varepsilon$$. So far so good.

But, for $$\delta \geq 1$$, we have $$\delta^2 \geq \delta$$ The range of value for $$\delta$$ in this case that I can think of such that $$\eqref{eq1}$$ and $$\eqref{eq3}$$ could be reconciled is $$\left|x^2\sin\left(\frac{1}{x}\right)\right| < \delta \leq \varepsilon$$

But when I check on a graphing tool, regardless of whether $$\delta$$ is $$<1$$ or $$\geq 1$$, $$0<\delta\leq\varepsilon$$ holds.

What am I doing wrong?

EDIT:

Since $$0<\delta\leq\varepsilon$$ and from $$\eqref{eq2}$$, we get $$|x|<\delta \leq\varepsilon \implies |x|^2 < \delta^2 \leq \varepsilon^2$$

So, since $$\left|x^2\sin\left(\frac{1}{x}\right)\right| \leq |x^2| \\ \implies \left|x^2\sin\left(\frac{1}{x}\right)\right| < \delta ^2 \leq \varepsilon^2$$

So, can we now say $$\delta \leq \varepsilon$$, regardless of values that either take?

• Recall that $\delta$ MAY depend on $\varepsilon$. So GIVEN $\varepsilon>0$, just take $\delta=\sqrt{\varepsilon}$. Jun 26, 2021 at 16:41

You're almost there, and there are several ways to finish up. One thing you could do is what @Tito Eliatron suggests in his comment (this might be the most elegant way):

Let $$\epsilon > 0$$, set $$\delta = \sqrt{\epsilon}$$. Then, for $$|x|<\delta$$ we have

$$\begin{equation} \left|x^2 \sin{\left(\frac{1}{x}\right)}\right| \leq |x|^2 < \delta^2 = \sqrt{\epsilon}^2 = \epsilon. \end{equation}$$

But there is another way to "fix" your argument. As you have noted, if $$\epsilon \leq 1$$, you can choose $$\delta = \epsilon$$ and use the fact that in this case we have $$\delta^2 \leq \delta$$, and do the same thing as above. Then we deal with the case $$\epsilon > 1$$ separately. In this case, we can choose $$\delta = 1$$, and we have for $$|x| < \delta:$$

$$\begin{equation} \left|x^2 \sin{\left(\frac{1}{x}\right)}\right| \leq |x|^2 < \delta^2 = 1^2 = 1 < \epsilon. \end{equation}$$

Thus, we can set $$\delta = \min{(\epsilon, 1)}$$ and obtain the same result. I think Tito's suggestion is more beautiful since it avoids different cases, but both methods are valid.

In $$\DeclareMathOperator{\epsilon}{\varepsilon}\epsilon$$-$$\delta$$ proofs, $$\delta$$ is allowed to depend on $$\epsilon$$, but not vice versa. In your example, if $$\epsilon\le1$$, then we can set $$\delta=\epsilon$$, and things work out nicely. Often, weird things can happen for large values of $$\epsilon$$, and so as a kind of 'safety-net' we require that $$\delta$$ be at most a specific number. Here, we can set $$\delta=\min\left(\epsilon,1\right)$$ and things work out nicely. It also helps to work backwards. We need $$\left|x^2\sin\left(\frac{1}{x}\right)\right|<\epsilon$$ or $$\left|x^2\right|\left|\sin\left(\frac{1}{x}\right)\right|<\epsilon \, .$$ Now, if we can make $$|x^2|$$ be smaller than $$\epsilon$$, then it will be certainly be the case that $$|x^2|\left|\sin(1/x)\right|$$ will be smaller than $$\epsilon$$, as $$|\sin(1/x)|\le1$$. Moreover, if $$|x|<1$$, then $$|x^2|<|x|$$. So if $$|x|<\epsilon$$ and $$|x|<1$$, then $$|x^2|<|x|<\epsilon$$, which is what we need. So let $$\delta=\min(\epsilon,1)$$, which guarantees that $$|x|<1$$ and $$|x|<\epsilon$$.