So, since we’ve made the assumption that the limit probably doesn’t exist that means we need to find two different paths upon which the limit has different values.

In this case note that using the \(x\)-axis or \(y\)-axis will not work as either one will result in a division by zero issue. So, let’s start off using the path \(x = {y^3}\).

Along this path we get,

\[\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^2} - {y^6}}}{{x{y^3}}} = \mathop {\lim }\limits_{\left( {{y^3},y} \right) \to \left( {0,0} \right)} \frac{{{y^6} - {y^6}}}{{{y^3}{y^3}}} = \mathop {\lim }\limits_{\left( {{y^3},y} \right) \to \left( {0,0} \right)} 0 = 0\] Show Step 3

Now let’s try \(x = y\) for the second path.

\[\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^2} - {y^6}}}{{x{y^3}}} = \mathop {\lim }\limits_{\left( {y,y} \right) \to \left( {0,0} \right)} \frac{{{y^2} - {y^6}}}{{y{y^3}}} = \mathop {\lim }\limits_{\left( {y,y} \right) \to \left( {0,0} \right)} \frac{{1 - {y^4}}}{{{y^2}}} = \mathop {\lim }\limits_{\left( {y,y} \right) \to \left( {0,0} \right)} \left( {\frac{1}{{{y^2}}} - {y^2}} \right) = \infty \] Show Step 4

So, we have two different paths that give different values of the limit and so we now know that this limit does not exist.