Proof involving the precise definition of a limit: is this wrong?

Question

I'm having some trouble with a proof involving the precise definition of a limit. I need to do the following:

Show that there exists $\delta > 0$ such that:

$\left|f(x,y)-f(0,0)\right|<\epsilon$ whenever $\sqrt{ x^2 + y^2 }<\delta$

with $f(x,y)=\frac{y}{x^2 + 1}$ and $\epsilon=0.05$

As you can see, this involves the precise definition of multivariable limits. In this particular case, we need to show that there exists $\delta > 0$ such that:

$\left|\frac{y}{x^2 + 1}\right| < 0.05$ whenever $\sqrt{ x^2 + y^2 } < \delta$

By the properties of absolute value, we have:

$\left|\frac{y}{x^2 + 1}\right|=\frac{|y|}{|x^2 + 1|}$

(note that $x^2 + 1\neq0$)

Then, we observe that:

$|y|\leq\sqrt{ x^2 + y^2 }$

and:

$|x^2+1|\geq1$

From this, it follows that:

$\frac{|y|}{|x^2 + 1|}\leq\sqrt{ x^2 + y^2 }$

Let $\delta=\epsilon=0.05$, then:

$\frac{|y|}{|x^2 + 1|}\leq\sqrt{ x^2 + y^2 }<\delta=\epsilon=0.05$

Thus, we have shown that for $\delta=0.05$:

$\left|\frac{y}{x^2 + 1}\right| < 0.05$ whenever $\sqrt{ x^2 + y^2 } < \delta$

Is my proof wrong? Are there alternative ways to do this? Do I have to take $\delta=0.05$ or can I take any $\delta$ that satisfies $0<\delta\leq\epsilon$?

Answer 1

I agree with the comments followng the question, which validates the OP's approach and therefore answers his question. Therefore, it's open season on an alternative approach.

I will derive a candidate specification between $\delta$ and $\epsilon$, and then verify it. Since I will be verifying at the end, I don't have to worry about whether each step that I take has a two-way implication (e.g. if and only if, rather than implies).

With it assumed that $\sqrt{x^2 + y^2} < \delta$ you know that $x^2 + y^2 < \delta^2.$
This means that $x^2 < \delta^2$ and $y^2 < \delta^2.$
This implies that $|x| < \delta$ and $|y| < \delta$.

Therefore, you have that $-\delta < x,y < \delta.$

Examining $|f(x,y)|$ you see that its denominator is $> 1$ and its numerator is $|y|.$

Further, since $f(0,0) = 0$ you have that $|f(x,y) - f(0,0)| = |f(x,y)| < |y| < \delta.$

Therefore, since it is desired that $|f(x,y) - f(0,0)| < \epsilon$,
the candidate specification is that $\delta = \epsilon.$

Verifying:

$\sqrt{x^2 + y^2} < \delta = \epsilon \implies x^2, y^2 < \epsilon^2.$

In particular, this implies that $|y| < \epsilon.$

Then, $|f(x,y) - f(0,0)| = |f(x,y)| = \frac{|y|}{x^2 + 1} < \frac{|y|}{1} = |y| < \epsilon.$

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Notes:

1. For a discussion of why I think that the Verification process is necessary see (specifically), the Addendum to this answer.

2. I think that the priority should be to make life as easy as possible on the problem solver, rather than trying to determine the tightest possible bound between $\epsilon$ and $\delta$. For example, suppose that I wanted to ensure both that $7\delta < \epsilon$ and that $5\delta^2 < \delta.$
   
   Then, I could simply specify $\delta = \min[(\epsilon/10), (1/10)].$

3. In the given problem, noticing that $|y| < \delta$ didn't really accomplish much. Suppose instead, that $f(x,y,z) = xy^2 + 3xyz$, and that you are examining $|(x,y,z) - (0,0,0)| < \delta$, in $\Bbb{R^3}.$
   
   Then, you can assume that each of $|x|,|y|,|z|$ must be less than $\delta$. This implies that $|f(x,y,z)| < 4\delta^3.$
   
   Therefore, if you artificially add the constraint that $\delta \leq (1/10)$, you are guaranteed that $4\delta^3 < \delta.$
   
   This facilitates having a linear constraint between $\delta$ and $\epsilon$, which greatly simplifies the verification process.