$\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(x)\sin(y)}{x^2+y^2}$ 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.
- Does this make sense?
- Does this method (the '3 steps') always work for functions of two variables?
- 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!
$\endgroup$ 51 Answer
$\begingroup$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 . Read two comments below of it and take a great way for such these limits. I hope it helps you.
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