Answered: x²y² lim(z,y)→(0,0) z²y²+(x−y)²

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Use two different paths to demonstrate that the limit does not exist.

Expert Solution

Step 1: Proving that the limit does not exist

$\lim_{(x, y) \to (0, 0)} \frac{x^{2} y^{2}}{x^{2} y^{2} + {(x - y)}^{2}}$

Let $y = m x$

$\lim_{(x) \to 0} \frac{x^{2} {(m x)}^{2}}{x^{2} {(m x)}^{2} + {(x - m x)}^{2}} = \lim_{(x) \to 0} \frac{x^{4} {(m)}^{2}}{x^{4} ({(m)}^{2} + {(1 - m)}^{2})} = \lim_{(x) \to 0} \frac{{(m)}^{2}}{({(m)}^{2} + {(1 - m)}^{2})} = \frac{m^{2}}{m^{2} + {(1 - m)}^{2}}$

Thus we can construct two function say $ϕ_{1} (x) = m_{1} x$ and $ϕ_{2} (x) = m_{2} x$ such that the limit of $f (x, ϕ_{1} (x))$ and $f (x, ϕ_{2} (x))$ are different, then the simultaneous limit doesn't exist

Similarly if we take $x = c y$

$\lim_{(x, y) \to (0, 0)} \frac{x^{2} y^{2}}{x^{2} y^{2} + {(x - y)}^{2}} = \lim_{y \to 0} \frac{{(c y)}^{2} y^{2}}{{(c y)}^{2} y^{2} + {(c y - y)}^{2}} = \frac{c^{2}}{c^{2} + {(c - 1)}^{2}}$

Thus we can construct two function say $φ_{1} (y) = c_{1} y$ and $φ_{2} (y) = c_{2} y$ such that the limit of $f (φ_{1} (y), y)$ and $f (φ_{2} (y), y)$ are different, then the simultaneous limit doesn't exist