Prove that $\lim \limits_{(x,y)\to (c,0)} \frac{\sin(x^2y)}{x^2-y^2}=0$, $c\ne0$ with (ε, δ)-definition

Question

So the problem is to work with $\left|\frac{\sin(x^2y)}{x^2-y^2}\right|$ and show that that is less than a formula involving δ, let´s call it g(δ), formula which I will later equal to ε in order to find the δs that guarantee: $||(x-c,y)||<δ⟹\left|\frac{\sin(x^2y)}{x^2-y^2}\right|<ε$

I have already researched extensively both StackExchange and the web in general, only to find that the closest examples to my limit formula out there have $x^2+y^2$ in the denominator, and the '+' there in place of the '-' appears to make a huge difference for cancelling purposes in the process of bounding.

I cannot get rid of, or should I the say 'control' the denominator no matter the approach I start with.

Below I itemize all the 'tricks' I have in my toolkit for the purpose of this proof, many of which I actually found trying to do it:

• $||(x-c,y)||<δ$
• $|x-c|<δ$ and $|y|<δ$
• Given that $(x-c)^2=x^2-2cx+c^2$, if I take $δ<|c|$, then (using 1st item) I get $x^2+y^2<2cx$
• $sign(x)=sign(c)$, or $2xc>0$ (it follows from last item)
• $|\sin(x^2y)|\le1$
• $|\sin(x^2y)|\le|x^2y|$
• $x^2-y^2=(x-y)(x+y)$

What do you think?

Answer 1

$\|(x-c,y)\| \le \delta$ implies $|x-c|\le \delta$ and $|y| \le \delta$.

Let $\delta \le |c|$ so that $|x-c| \le \delta$ implies $|x| \le |x-c|+|c| \le 2|c|$. Using one of your observations, we have $|\sin(x^2 y )| \le |x^2 y| \le (2c)^2 |y| \le 4c^2 \delta$.

Let $\delta \le |c|/2$ so that $|x-c| \le \delta$ implies $|x| \ge |c| - |x-c| \ge |c|/2$. Then, $|x^2-y^2| \ge x^2 - y^2 \ge c^2/4 - \delta^2$. If we further let $\delta \le |c|/4$, then we have $c^2/4-\delta^2 \ge 3c^2/16$

Combining these two inequalities, we have that for $\delta \le |c|/4$, we have $\left|\frac{\sin(x^2 y)}{x^2-y^2}\right| \le \frac{4c^2 \delta}{3c^2/16} = 64\delta/3$. Then you can choose $\delta = 3\epsilon/128$ to make this upper bound less than $\epsilon$. In short, choosing $\delta = \min\{3\epsilon/128, |c|/4\}$ works.

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The intuition is as follows: $x^2 y \to 0$, so the numerator gets close to $0$. The denominator $x^2-y^2 \to c^2$, and in particular it stays bounded away from zero (e.g., it eventually stays larger than $3c^2/16$). So, the ratio tends to zero.

Answer 2

First, note that

$$\left|\frac{\sin(x^2y)}{x^2-y^2}\right|\leq\left|\frac{\sin(x^2y)}{x^2}\right|$$

because the denominator can only increase in size from the right hand side to the left hand side.

Now if $||(x-c, y)||<\left|\frac{c}{2}\right|$ then $|x|\leq\left|\frac{c}{2}\right|$, so

$$\left|\frac{\sin(x^2y)}{x^2}\right|\leq\left|\frac{\sin(x^2y)}{\left|\frac{c}{2}\right|^2}\right|=\frac{\left|\sin(x^2y)\right|}{\left(\frac{c}{2}\right)^2},$$

so it is enough to show that there exists a $\delta$ such that

\begin{equation} ||(x-c,y)||<\delta\Rightarrow\frac{\left|\sin(x^2y)\right|}{\left(\frac{c}{2}\right)^2}<\varepsilon \tag{1} \end{equation}

(If the $\delta$ we find that makes this true is larger than $\left|\frac{c}{2}\right|$ then we can just take $\min(\delta, \left|\frac{c}{2}\right|)$).

$(1)$ is true if and only if

\begin{equation} ||(x-c,y)||<\delta\Rightarrow\left|\sin(x^2y)\right|<\frac{4\varepsilon}{c^2}. \tag{2} \end{equation}

Do I need to continue?