Section 13.1 : Limits
4. Evaluate the following limit.
\[\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^2} - {y^6}}}{{x{y^3}}}\]Show All Steps Hide All Steps
Start SolutionOkay, with this problem we can see that, if we plug in the point, we get zero in the numerator and the denominator. Unlike the second problem above however there is no factoring that can be done to make this into a “doable” limit.
Therefore, we will proceed with the problem as if the limit doesn’t exist.
Show Step 2So, 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 3Now 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 4So, we have two different paths that give different values of the limit and so we now know that this limit does not exist.