Question

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

\(\mathop {lim}\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{5{y^4}co{s^2}x}}{{{x^4} + {y^4}}}\)

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 \(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{5{y^4}{{\cos }^2}x}}{{{x^4} + {y^4}}}\)

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

\(\begin{aligned}{l}f(x,y) = \frac{{5{y^4}{{\cos }^2}x}}{{{x^4} + {y^4}}}\\f(x,0) = \frac{{5{{(0)}^4}{{\cos }^2}x}}{{{x^4} + {{(0)}^4}}}\\f(x,0) = \frac{0}{{{x^4}}}\\f(x,0) = 0\end{aligned}\)

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

\(\begin{aligned}{l}f(x,y) = \frac{{5{y^4}{{\cos }^2}x}}{{{x^4} + {y^4}}}\\f(0,y) = \frac{{5{y^4}{{\cos }^2}(0)}}{{{{(0)}^4} + {y^4}}}\end{aligned}\)

As\(cos0 = 1\)

\(\begin{aligned}{l}f(0,y) = \frac{{5{y^4}{{(1)}^2}}}{{{y^4}}}\\f(0,y) = 5\end{aligned}\)

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 \(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{5{y^4}{{\cos }^2}x}}{{{x^4} + {y^4}}}\) does not exist.