Prove that the limit does not exist: $\lim_{(x,y)\to(1,3)}\frac{|\sin(3x-y)|}{\sqrt{2(x-1)^2+5(y-3)^4}}$ (Without L'Hopital's)

Question

It seems that the function is underfined at $(1,3)$, so when I try to directly plug in $(1,3)$ via limit laws, I get an undetermined term $\frac{0}{0}$. So I tried instead to consider individual paths in the domain leading towards $(1,3)$, i.e.,

Along the x-axis: fixing $y=0$, $$f(x,0)=\frac{|\sin(3x)|}{\sqrt{2(x-1)^2+5(y-3)^4}}$$

However, I don't see where to go from there (if this is even the right approach), and try to show that two different paths that I pick result in differing limits.

For these type of questions in general, am I supposed to simply "try" a bunch of paths towards my limit point $(a,b)$ and simply "hope" that two paths will give different limits, or can we do better with a faster method?

Answer 1

Consider the paths $y=mx+3-m$ for $m\in\Bbb R^+$ passing through the point $(1,3)$. Along these paths, the given limit becomes $$\lim_{(x,y)\to(1,3)}\frac{|\sin(3x-y)|}{\sqrt{2(x-1)^2+5(y-3)^4}}=\lim_{x\to 1}\left|\frac{\sin((m-3)(x-1))|}{x-1}\right|\frac1{\sqrt{2+5m^4(x-1)^2}}$$ which is equal to $\frac{|m-3|}{\sqrt2}.$ It depends on $m$. So, there is no limit. Note that the trigonometric limit is well-known and it can be calculated without L'Hopital's rule.

Answer 2

First reduce to $(u,v)→(0,0)$ by letting $x=1+u,y=3+v.$ Then, the first paths to try are the linear ones, $v=mu.$ If your limit depends on $m,$ you have won. Else, try more "ad hoc" paths.

Your last comment confirms that you completely understood: as $u\to0,$ $$\begin{align}\frac{|\sin(3x-y)|}{\sqrt{2(x-1)^2+5(y-3)^4}}&=\frac{|\sin(3u-v)|}{\sqrt{2u^2+5v^4}}\\&=\frac{|\sin((3-m)u)|}{\sqrt{2u^2+5m^2u^4}}\\&\sim\frac{|3-m||u|}{|u|\sqrt{2+5m^2u^2}}\\&\to\frac{|3-m|}{\sqrt2}\end{align}$$ which depends on $m,$ thereby proving that $\lim_{(x,y)\to(1,3)}\frac{|\sin(3x-y)|}{\sqrt{2(x-1)^2+5(y-3)^4}}$ does not exist.

You might also notice that the limit along the line $u=0$ (as $v\to0$) is $+\infty.$