Question

Find the limit, if it exists, or show that the limit does not exist.

\(\mathop {lim}\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{x{y^4}}}{{{x^2} + {y^8}}}\)

Short answer

The limit of the given function does not exist.

Step-by-step solution

Check whether the limit exist or not

Consider the function \(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{x{y^4}}}{{{x^2} + {y^8}}}\)

Let’s approach\(\left( {0,0} \right)\)along the X-axis. i.e. \(y = 0\) and then find the limit as\(x\)approaches to\(0\).

\(\begin{aligned}{l}f(x,y) = \frac{{x{y^4}}}{{{x^2} + {y^8}}}\\f(x,0) = \frac{{x{{(0)}^4}}}{{{x^2} + {{(0)}^8}}}\\f(x,0) = 0\end{aligned}\)

Let’s approach\(\left( {0,0} \right)\)along the Y-axis. i.e. \(x = 0\) and then find the limit as\(y\)approaches to\(0\).

\(\begin{aligned}{l}f(x,y) = \frac{{x{y^4}}}{{{x^2} + {y^8}}}\\f(0,y) = \frac{{(0){y^4}}}{{{{(0)}^2} + {y^8}}}\\f(0,y) = 0\end{aligned}\)

Let’s approach\(f(x,y)\)along the curve \(x = {y^4}\) and passes through\((0,0)\) as required.

\(\begin{aligned}{l}f(x,y) = \frac{{x{y^4}}}{{{x^2} + {y^8}}}\\f({y^4},y) = \frac{{{y^4}{y^4}}}{{{{({y^4})}^2} + {y^8}}}\\f({y^4},y) = \frac{{{y^8}}}{{2{y^8}}}\end{aligned}\)

Taking the limit,

\(\begin{aligned}{l}\mathop {\lim }\limits_{y \to 0} f({y^4},y) = \mathop {\lim }\limits_{y \to 0} \frac{{{y^8}}}{{2{y^8}}}\\\mathop {\lim }\limits_{y \to 0} f({y^4},y) = \frac{1}{2}\end{aligned}\)

Compare the both limits

The given function f has two different limits along two different lines, the limit does not exist.

Hence, the limit of the given function \(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{x{y^4}}}{{{x^2} + {y^8}}}\) does not exist.