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Found in: Page 632

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Use a computer graph of the function to explain why the limit does not exist.$$\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}}$$

The limit of function along different paths y = x and $$y = {x^{\frac{1}{3}}}$$ is different.

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

See the step by step solution

## Step 1: Using Computer Software to Draw a Graph:

Consider the following limit,

$$\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}}$$

The graph of the function $$z = f\left( {x,{\rm{ }}y} \right) = \frac{{x{y^3}}}{{{x^2} + {y^6}}}$$ in the window

\begin{aligned}{l}\left( {0,{\rm{ }}0.001} \right) \times \left( {0.01} \right) \times \left( {0.05} \right) \to {\rm{ Graph using computer software}}\\{\rm{which gives }}z = 0.5{\rm{ where }}y = {x^{\frac{1}{3}}}\end{aligned}

## Step 2: Limit of Function:

\begin{aligned}{l}\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x{{({x^{\frac{1}{3}}})}^3}}}{{{x^2} + {{({x^{\frac{1}{3}}})}^6}}}\\{\rm{ }} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{{x^2} + {x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{2{x^2}}} = \frac{1}{2} = 0.5\\\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}} = 0.5\end{aligned}

## Step 3: Graph of $$\frac{{x{y^3}}}{{{x^2} + {y^6}}}$$

\begin{aligned}{l}(0,0.0001) \times (0,0.00001) \times (0,4 \times {10^{ - 4}}){\rm{ using any computer software gives the graph}}{\rm{.}}\\\\{\rm{From the graph along the line }}y = x{\rm{ as }}(x,y) \to (0,0){\rm{ the function is constant }}z = 0\end{aligned}

The limit of this function is 0

\begin{aligned}{l}\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x{{(x)}^3}}}{{{x^2} + {{(x)}^6}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^4}}}{{{x^2} + {x^6}}}\\{\rm{ }} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{1 + {x^4}}}\\{\rm{ }} = \frac{0}{{1 + 0}} = 0\\\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}} = 0\\\\{\rm{The limit of this function is not independent of path passing through (0,0)}}{\rm{. And }}(x,y) \to (0,0)\\{\rm{The limit of function along different paths }}y = x{\rm{ and }}y = {x^{\frac{1}{3}}}{\rm{ is different}}{\rm{.}}\end{aligned}

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