Question

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

\(\mathop {lim}\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \left( {5{x^3} - {x^2}{y^2}} \right)\)

Short answer

The limit of the given function exists.

The value of the limit of the function is: \(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \left( {5{x^3} - {x^2}{y^2}} \right) = 1\).

Step-by-step solution

Check whether the limit exist or not

Consider the function \(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \left( {5{x^3} - {x^2}{y^2}} \right)\)

The function\(f(x,y) = \left( {5{x^3} - {x^2}{y^2}} \right)\) is polynomial. So it is continuous at everywhere.

So the limit is exists.

Find the limit

\(\begin{aligned}{l}\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \left( {5{x^3} - {x^2}{y^2}} \right) = f(1,2)\\\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \left( {5{x^3} - {x^2}{y^2}} \right) = 5{(1)^3} - {(1)^2}{(2)^2}\\\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \left( {5{x^3} - {x^2}{y^2}} \right) = 1\end{aligned}\)

Therefore, the given function is continuous. The limit is exists. The value of the limit of the function is: \(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \left( {5{x^3} - {x^2}{y^2}} \right) = 1\).