A 2-Part Test (or The Squeeze Thm.) Question

Question

Find the limit, if it exists.

I am sure that one of the paths which uses the line y=mx does not work.

I am not sure for y=x^m because it lead me to *mtan(x^2+m^2*x^2)/(x^2+m^2)*

What should I do?

]1

Answer 1

Let consider the more general

$$\lim_{x,y\to0} \frac{xy \tan(x^2+y^2)}{x^4+y^2}$$

we have that

$$\frac{xy \tan(x^2+y^2)}{x^4+y^2}=\frac{xy(x^2+y^2)}{x^4+y^2}\frac{ \tan(x^2+y^2)}{x^2+y^2}\to 0$$

indeed

$$\frac{ \tan(x^2+y^2)}{x^2+y^2}\to 1$$

and

$$\frac{xy(x^2+y^2)}{x^4+y^2}\to0$$

indeed consider

$$\frac{|xy|(x^2+y^2)}{x^4+y^2}$$

and let

• $x^2=r \cos \theta$
• $y=r \sin \theta$

thus

$$0\le\frac{|xy|(x^2+y^2)}{x^4+y^2}=\frac{r^3\,\sqrt r\sqrt{|\cos \theta|}\sin \theta}{r^2}=r\,\sqrt r\sqrt{|\cos \theta|}\sin \theta \to0$$

and for squeeze theorem

$$\frac{|xy|(x^2+y^2)}{x^4+y^2}\to 0 \implies \frac{xy(x^2+y^2)}{x^4+y^2}\to0$$