3
$\begingroup$

As the title indicates, Using $\epsilon$-$\delta$ approach to prove that $$\lim_{(x,y,z)\rightarrow (0,0,0)}\frac {y^3-1000xy^2+z^5}{x^2+y^2+z^4}=0$$?

$\endgroup$
8
$\begingroup$

Here is a method without spherical coordinates, only crude inequalities. First, we have:

$$\left| \frac{y^3-1000xy^2+z^5}{x^2+y^2+z^4} \right| \leq \frac{|y|^3}{x^2+y^2+z^4}+1000\frac{|x|y^2}{x^2+y^2+z^4}+\frac{|z|^5}{x^2+y^2+z^4}.$$

Since each of $x^2$, $y^2$ and $z^4$ is non-negative, we have $x^2+y^2+z^4 \geq y^2$ and $x^2+y^2+z^4 \geq z^4$, so that:

$$\left| \frac{y^3-1000xy^2+z^5}{x^2+y^2+z^4} \right| \leq \frac{|y|^3}{y^2}+1000\frac{|x|y^2}{y^2}+\frac{|z|^5}{z^4} = |y|+1000|x|+|z|.$$

Hence, the limit at $(0,0,0)$ of this function is $0$.

$\endgroup$
1
$\begingroup$

Let $$ \begin{eqnarray} x &=& r\cos\phi\cos\theta &=& rau &\qquad& a &=& \cos\phi &\qquad& u &=& \cos\theta \\ y &=& r\cos\phi\sin\theta &=& rav &\qquad& b &=& \sin\phi &\qquad& v &=& \sin\theta \\ z &=& r\sin\phi &=& rb \end{eqnarray} $$ Then we have $$ \lim_{r \to 0} \frac{r^3a^3v^2(v-1000u)+r^5b^5}{r^2a^2+r^4b^4} = \lim_{r \to 0} \frac{r a^3v^2(v-1000u)+r^3b^5}{ a^2+r^2b^4} = 0 $$ which vanishes everywhere (and is thus well-defined). To see this, consider first that when $a\neq0$, the denominator is dominated by the $a^2$ term so that the limit becomes $$ \lim_{r \to 0} \; rav^2(v-1000u) = 0 $$ while when $a=0$ (i.e. at the north and south poles where $b=\pm1$), the limit becomes $$ \lim_{r \to 0} \frac{r^3b^5}{r^2b^4} = \lim_{r \to 0} \; rb = 0 $$ which is still independent of the direction of approach to the origin.

$\endgroup$
  • $\begingroup$ I don't understand why you write the last equality: $\lim_{r\to 0}\frac{r^3b^5}{r^2b^4}=\lim_{r\to 0} r\sin\phi=0$... $\endgroup$ – Fosco Jan 23 '12 at 10:46
  • $\begingroup$ Thank you @tetrapharmakon, i fixed my algebraic error, and the limit is $0$. $\endgroup$ – bgins Jan 23 '12 at 12:13
0
$\begingroup$

If you choose spherical coordinates and fiddle with trigonometric inequalities you should obtain 0... Am I missing something?

$\endgroup$
  • $\begingroup$ is there any other apporach to show its limit exists other than using spherial coordination system. $\endgroup$ – Mathematics Jan 23 '12 at 10:09
  • $\begingroup$ I think coordinate-change is the neatest way to prove/disprove that limit exists... Have you been assigned this as an homework? Is it linked to a "real-life" problem you are facing? What's the approach you are used to see in solving this kind of problems? $\endgroup$ – Fosco Jan 23 '12 at 10:50
  • $\begingroup$ @Mathematics Can you modify the OP in order to answer to my question? $\endgroup$ – Fosco Jan 23 '12 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.