Multi-variable Limit

Analysis is the study of Functions, using the main tool "Limit".

Limit problems appear in:
1. Continuity
2. Derivative
3. Integral
4. Sequence

Multi-variable Functions have different approach of Limit compared to Single-value functions.
Eg. L'Hôpital Rule is not applicable to Multi-variable Functions.

Case 1: Find the Limit (L) of
$latex \displaystyle f(x,y)= \frac{xy}{x^{2}+y^{2}} \text{ at point (0,0)} $

Solution:
Consider the point P(x,y) on f(x,y)
$latex \displaystyle \lim_{x\to 0} f(x,y)=f(0,y)= 0 \text{...(1)}$
$latex \displaystyle\lim_{y\to 0} f(x,y)=f(x,0)=0 \text{...(2)}$

but when P moves along y=x straight line approaching (0,0),
ie. x->0, y=x->0,
$latex \displaystyle\lim_{x\to 0} \frac{xy}{x^{2}+y^{2}}= \frac{x^2}{2.x^2} = \frac{1}{2} \text{...(3)}$

From (1),(2),(3) there are 3 limits {0, 0, 1/2}, hence the Limit L does not exist.

Case 2:
Find limit L of
$latex \displaystyle f(x,y)= \frac{x^{2}+y^{2}}{x^{4}+y^{4}}
\text { at point} (\infty, \infty)$

Solution:
Let y= kx
$latex \displaystyle f(x,y)= \frac{x^{2}+k^{2}y^{2}}{x^{4}+k^{4}y^{4}} = \frac{1+k^{2}}{(1+k^{4})x^2}$
When x -> $latex \infty$,
f(x,y) independent of k => possible limit of 0

Prove: $latex \displaystyle\lim_{(x,y)\to \infty} f(x,y)= L = 0$

$latex \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| = \frac{x^{2}+y^{2}}{x^{4}+y^{4}}
\leq \frac{x^{2}+y^{2}}{2x^{2}y^{2}} = \frac{1}{2x^{2}} + \frac{1}{2y^{2}}$

Note:
$latex (x^{2} - y^{2})^2 \geq 0$
$latex x^{4}-2x^{2}y^{2}+y^{4}\geq 0 $
$latex x^{4}+y^{4} \geq 2x^{2}y^{2}$

$latex \displaystyle\forall \epsilon > 0 \text{, take } \delta \geq \frac{1}{\sqrt{\epsilon}} \text{ such that}$
$latex |x|>\delta, |y|>\delta$
$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \frac{1}{2x^{2}} + \frac{1}{2y^{2}}
< \frac{1}{2{\delta}^{2}}+\frac{1}{2{\delta}^{2}}$

$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \frac{1}{2}\left({\frac{1}{\sqrt{\epsilon}}}\right)^{-2}+\frac{1}{2}\left({\frac{1}{\sqrt{\epsilon}}}\right)^{-2} = \frac{\epsilon}{2}+ \frac{\epsilon} {2}$
$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \epsilon$
$latex \implies \displaystyle\lim_{(x,y)\to \infty} f(x,y)= L =0 $
[QED]

Source: Prof Zhang ShiZao 章士藻(1940-) "Collected Works of Mathematics Education" 数学教育文集
for Ecole Normale Supérieure in China 高等师范学院