Finding limit $ \lim_{{x\rightarrow0},y\rightarrow0}\frac{\cos xy-2^{x{y}^2}}{x^2y^2}$ Compute or show that a limit
$$L: \lim_{(x,y)\to(0,0)}\frac{\cos xy-2^{x{y}^2}}{x^2y^2}$$
doesn't exist. Here is my approach:
First I divided the limit into two limits first one is
$\lim_{{x\rightarrow0},y\rightarrow0}\frac{\cos xy}{x^2y^2}$ (Lets call this limit $L_1$)
And second is $\lim_{{x\rightarrow0},y\rightarrow0}\frac{2^{x{y}^2}}{x^2y^2}$. (And call this $L_2$)
When solving $L_1$ I substituted $xy=t, t\rightarrow 0$ and I get
$\lim_{{t\rightarrow0}}\frac{\cos t}{t^2}$, Since this is $\frac{constant}{0}$ this is undefined end hence the limit does not exist, which indicates that also limit $L$ does not exist?
Is this okay, if not what other approach can be made?

To see explicitely that the limit doesn't exist let consider $x=y= t \to 0$
$$\frac{\cos xy-2^{x{y}^2}}{x^2y^2}=\frac{\cos t^2-2^{t^3}}{t^4}=\frac{\cos t^2-1}{t^4}-\frac1{t}\frac{2^{t^3}-1}{t^3}\to -\frac12\pm\infty\cdot\log 2=\pm\infty$$
