Two variable limits

[dunkelheit]

Evaluate, if it exists, the limit

[MATH]\lim_{(x,y)\to(0,0)} \frac{\sin(xy^2)}{(x^2+y^4)^{\frac{1}{2}}}[/MATH]​

My try: since [MATH]|\sin t| \leq |t|[/MATH] for all [MATH]t\in\mathbb{R}[/MATH] and for (increasing) monotonicity of the square root and (decreasing) monotonicity of the reciprocal it is

[MATH]x^2+y^4 \geq y^4 \geq 0 \Longleftrightarrow (x^2+y^4)^{\frac{1}{2}} \geq (y^4)^{\frac{1}{2}}=|y^2|=y^2 \Longleftrightarrow \frac{1}{(x^2+y^4)^{\frac{1}{2}}} \leq \frac{1}{y^2} \Longleftrightarrow \frac{|\sin(xy^2)|}{(x^2+y^4)^{\frac{1}{2}}} \leq \frac{|\sin(xy^2)|}{y^2} \leq \frac{|xy^2|}{y^2}=\frac{|x||y^2|}{y^2}=|x| [/MATH]​

So since the limit preserves the non strict inequalities and for the continuity of the modulus it is

[MATH]0 \leq \left|\lim_{(x,y)\to(0,0)} \frac{\sin(xy^2)}{(x^2+y^4)^{\frac{1}{2}}}\right|=\lim_{(x,y)\to(0,0)} \left|\frac{\sin(xy^2)}{(x^2+y^4)^{\frac{1}{2}}}\right|=\lim_{(x,y)\to(0,0)} \frac{|\sin(xy^2)|}{(x^2+y^4)^{\frac{1}{2}}} \leq \lim_{(x,y)\to(0,0)} |x|[/MATH]​

Since [MATH]|x|[/MATH] is a continuous function, its limit as [MATH](x,y)\to(0,0)[/MATH] can be evaluated by substitution; so

[MATH]0 \leq \lim_{(x,y)\to(0,0)} \left|\frac{\sin(xy^2)}{(x^2+y^4)^{\frac{1}{2}}}\right| \leq \lim_{(x,y) \to (0,0)} |x| =0[/MATH]​

So by the squeeze theorem it follows that

[MATH]\left|\lim_{(x,y)\to(0,0)} \frac{\sin(xy^2)}{(x^2+y^4)^{\frac{1}{2}}}\right|=0[/MATH]​

Since in [MATH]\mathbb{R}[/MATH] it is [MATH]|v|=0 \Longleftrightarrow v=0[/MATH], this implies that

[MATH]\lim_{(x,y)\to(0,0)} \frac{\sin(xy^2)}{(x^2+y^4)^{\frac{1}{2}}}=0[/MATH]​

Is this correct? If there is any mistake or imprecision in what I've said, can someone please stress it? Another question: WolframAlpha says that the limit doesn't exist, but it specifies "value may depend on [MATH]x,y[/MATH] path in complex space", so have I done some mistakes or the "complex space" has different rules for limits? Thanks.

 

[LCKurtz]

I see nothing wrong with your reasoning. I guess if someone were to be super pedantic about it, they might ask how you know that [MATH]|x|\to 0[/MATH] follows from [MATH](x,y)\to (0,0)[/MATH].

 

[dunkelheit]

Thanks for your confirmation! Maybe because [MATH]0 \leq |x|=\sqrt{x^2} \leq \sqrt{x^2+y^2}=\|(x,y)\|[/MATH] and to show that [MATH]|x|\to0[/MATH] as [MATH](x,y)\to(0,0)[/MATH] means showing that for all [MATH]\varepsilon>0[/MATH] there exists [MATH]r>0[/MATH] such that [MATH]\|(x,y)\|<r[/MATH] implies [MATH]||x|-0| < \varepsilon \Longleftrightarrow |x| <\varepsilon[/MATH].
But [MATH]0 \leq |x| \leq \|(x,y)\|[/MATH], so it is sufficient to take [MATH]r<\varepsilon[/MATH] and we get [MATH]0 \leq |x| \leq \|(x,y)\|<r< \varepsilon \implies |x| < \varepsilon[/MATH]. Is this correct?

 

[LCKurtz]

Yes.