2
$\begingroup$

Show that

$$\lim_{(x,y)\to (0,0)}\frac{x^2-y^2}{x^2+y^2}\sin(x-3y)$$

Does not exists

I've tried the traditional patches, but I always find zero as answer. Any hint?

Thanks in advance!

$\endgroup$
6
$\begingroup$

Note that

$$\left|\frac{x^2-y^2}{x^2+y^2}\right|\le 1.$$ So

$$\left|\frac{x^2-y^2}{x^2+y^2}\sin(x-3y)\right|\le |\sin(x-3y)|.$$

Moreover

$$\lim_{(x,y)\to (0,0)}\sin(x-3y)=0.$$ Thus

$$\lim_{(x,y)\to (0,0)}\left| \frac{x^2-y^2}{x^2+y^2}\sin(x-3y)\right|\le \lim_{(x,y)\to (0,0)}|\sin(x-3y)|=0,$$ from where we conclude that the original limit exists and its value is $0.$

$\endgroup$

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.