Question

Use a computer graph of the function to explain why the limit does not exist.

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}}\)

Short answer

The limit of the function along paths (y = x and y = – x) is different.

The limit of this function does not exist at the point (x, y) = (0, 0)

Step-by-step solution

Step 1: Graph of the Function:

Consider the following limit

\(\begin{aligned}{l}\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}}\\\\{\rm{The graph of the function }}z = f(x,y) = \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}}{\rm{ in the window }}\\(0,0.0001) \times (0,0.00001) \times (0.6,1.125){\rm{ is shown below}}{\rm{. (Use any computer software for graph)}}\end{aligned}\)

Step 2: Calculating Limit:

From the graph, along the line y = x as (x, y)\( \to \)(0, 0), that is constant z = 1.125

The limit of this function is 1.125

\(\begin{aligned}{l}\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{x^2} + 3xx + 4{x^3}}}{{3{x^2} + 5{x^2}}}\\{\rm{ }} = \mathop {\lim }\limits_{x \to 0} \frac{{9{x^2}}}{{8{x^2}}} = \frac{9}{8}\\\\\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}} = 1.125\end{aligned}\)

Step 3: Graph when y = – x

\(\begin{aligned}{l}{\rm{The graph of the function }}z = f(x,y) = \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}}{\rm{ in the window: }}\\(0,0.0001) \times ( - 0.00001,0) \times (0.375,0.8){\rm{ }}\end{aligned}\)

Step 4: Limit of this Function:

From the graph, along the line y = – x as (x, y)\( \to \)(0, 0), that is constant z = 0.375

The limit of this function is 0.375

\(\begin{aligned}{l}\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{x^2} - 3xx + 4{x^2}}}{{3{x^2} + 5{x^2}}}\\{\rm{ }} = \mathop {\lim }\limits_{x \to 0} \frac{{3{x^2}}}{{8{x^2}}} = \frac{3}{8}\\\\\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}} = 0.375\end{aligned}\)

The limit of this function is not independent of the path passing through (0, 0)

And (x, y)\( \to \)(0, 0)

Hence, the limit of function along different paths (y = x and y = – x) is different.

The limit of this function does not exist at the point (x, y) = (0, 0)