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

Question

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?

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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?

Answer 1

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.

Answer 2

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 $|x|<\epsilon$.