0
$\begingroup$

$\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(x)\sin(y)}{x^2+y^2}$ the limit is undefined.

I would like to know if the following method is sufficient to prove that the limit is undefined:

step1: $\lim_{(x,y)\rightarrow(x,0)}\frac{\sin(x)\sin(0)}{x^2+0^2}=0$

step2: $\lim_{(x,y)\rightarrow(0,y)}\frac{\sin(0)\sin(y)}{0^2+y^2}=0$

step3: but $\lim_{(x,y)\rightarrow(x,x)}\frac{\sin(x)\sin(x)}{x^2+x^2}=\frac{\sin(x)^2}{2x^2}=\frac{1}{2}(\frac{\sin x}{x})^2$

and as $x$ then $\rightarrow 0$ $\frac{\sin(x)^2}{2x^2}=\frac{1}{2}(\frac{\sin x}{x})^2 \rightarrow \frac{1}{2}$

hence the limit does not exist.

  1. Does this make sense?
  2. Does this method (the '3 steps') always work for functions of two variables?
  3. Would it be valid to use the same method to prove: $\lim_{(x,y)\rightarrow(0,0)}\frac{x^3-y^3}{x^2+y^2}=0$

Thank you!

  • 0
    I have my answer for the example I solved above; but for the other example below: lim(x,y)--> (0,0) for {x^3-y^3}/{x^2+y^2}=0; I have no idea how to prove mathematically.2012-10-17

1 Answers 1

1

About the third: In this kinds of limits, you can take $r_α (t)=(t,αt)$ and put it into your function to find path wise limit of $f$ when $t$ tends to $0$ . After simplifying the original function, if the limit of last expression approaches to zero then probably your original function has limit $0$ at $(0,0)$ . Now, you have to use $ϵ,δ$ to prove your limit. Note that we see $\lim_ {t→0} r_α (t)=(0,0)$ . There are some small magic points in these limits also. See what @Jennifer Dylan here noted https://math.stackexchange.com/a/128564/8581. Read two comments below of it and take a great way for such these limits. I hope it helps you.

  • 0
    tha$n$ks! that $m$a$k$es se$n$se :)2012-10-17