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I came across the following problem as i was reading an proof for limit of $f(x, y) =\large \frac{4xy^2}{x^2+y^2}$ using epsilon delta definition of the limit.

The book is Thomas Calculus 14th edition, chapter 14.

Below is the line of proof i could not understand

$\Large\frac{4|x|y^2}{x^2+y^2}$ $\le 4|x| = 4\sqrt{x^2}\le 4\sqrt{x^2+y^2}$

After which they say if we let $\delta = \frac{\epsilon}{4}$

They lost me in the inequalities, help me understand in simple words and also tell me what is the entire point of this epsilon delta jargon.

MathematicianByMistake
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Anonymous
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    Is it that you do not understand how the inequality is derived? – 温泽海 Sep 12 '24 at 15:41
  • @cansomeonehelpmeout has provided an answer for why the inequality is true. Explaining "the entire point of all this epsilon delta jargon" is a much harder question to answer. The accepted answer here might help: https://math.stackexchange.com/questions/3658590/understanding-epsilon-delta-proof?rq=1 – Ethan Bolker Sep 12 '24 at 15:46
  • $x^2 \ge 0$ so $y^2 \le x^2+y^2$ so $\frac {y^2}{x^2 + y^2} \le 1$. Are you with me so far? So $\frac {4|x|y^2}{x^2+y^2} = 4|x|\cdot \frac{y^2}{x^2+y^2}=4|x|\cdot 1 = 4|x|$. Meanwhile $|x| = \sqrt{x^2}$ while $y^2\ge 0$ so $x^2 + y^2 \ge x^2$ and $\sqrt{x^2+y^2} \ge \sqrt{x^2} = |x|$. So $4|x|=4\sqrt{x^2} \le 4\sqrt{x^2 + y^2}$. – fleablood Sep 12 '24 at 15:56

2 Answers2

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If $y=0$ then $0\leq 1$ is true. If $y\neq 0$, then $$\frac{y^2}{x^2+y^2}=\frac{1}{\left(\frac{x}{y}\right)^2+1}\leq 1$$

cansomeonehelpmeout
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This all relies on $K^2 \ge 0$ so:

  • $K^2 + M^2 \ge M^2$
  • $\sqrt{K^2+M^2} \ge \sqrt{M^2}$
  • $\frac {M^2}{K^2 + M^2} \le 1$

The text is assuming the reader is fully comfortable with the conclusions and sees them immediately without question.

To add in extra steps:

$\frac {4|x|y^2}{x^2 + y^2} = $

$4|x|\cdot\frac {y^2}{x^2 + y^2} =$

$4|x|\cdot\frac {0 + y^2}{x^2 + y^2}\le$

$4|x|\cdot\frac {x^2 + y^2}{x^2 + y^2} =$

$4|x|\cdot 1=$

$4|x| = $

$4\sqrt{x^2} = $

$4\sqrt{x^2 + 0} \le $

$4\sqrt{x^2 + y^2}$

I really hope this now makes everything very clear.

fleablood
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