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Limits.

Definition 4.4   Let f be a function defined on a (pointed) neighborhood of P0=(x0,y0) and $l \in \mathbb{R} $. The function f has a limit equal to l when P=(x,y) is arbitrarily close to P0 if:

\begin{displaymath}\forall \varepsilon >0, \; \exists \delta >0 \vert d(P,P_0) < \delta \Longrightarrow \vert f(P)-l\vert < \varepsilon
\end{displaymath}

We denote:

\begin{displaymath}\underset{(x,y) \rightarrow (x_0,y_0)}{\text{lim}} f(x,y)= l ...
...xt{or}
\qquad \underset{P \rightarrow P_0}{\text{lim}} f(P)= l
\end{displaymath}

Proposition 4.5   If it exists, the limit of f at P0 is unique.

Example 4.6   Let f be the function defined by

\begin{displaymath}\forall (x,y) \neq (0,0), \; f(x,y) = \frac {x^2y}{x^4+y^2}
\end{displaymath}

Approach the origin on the parabola $\mathcal{P}_m$ whose equation is y=mx2, where m is a real parameter different from 0. On $\mathcal{P}_m$, we have:

\begin{displaymath}f(x,y)=f(x,mx^2)=\frac {mx^4}{x^4+m^2x^4} = \frac 1m
\end{displaymath}

Thus the limit at 0 of the function f, computed on the parabola $\mathcal{P}_m$ is equal to $\frac 1m$, i.e. on each such parabola, there is a separate ``candidate'' to be the limit of f at (0,0). Thus there is no limit for f at the origin.

Proposition 4.7 (Algebra of the limits)   Let f and g be two functions, and suppose that

\begin{displaymath}\underset{P \rightarrow P_0}{\text{lim}} f(P)= l_1 \qquad \text{and} \qquad
\underset{P \rightarrow P_0}{\text{lim}} g(P)= l_2
\end{displaymath}

Then:
1.
Sum: $\underset{P \rightarrow P_0}{\text{lim}} [f(P) \pm g(P)]= l_1 \pm l_2$.
2.
Product: $\underset{P \rightarrow P_0}{\text{lim}} [f(P)g(P)]= l_1 l_2$.
3.
Constant multiple: $\forall k \in \mathbb{R} , \;
\underset{P \rightarrow P_0}{\text{lim}} kf(P)= kl_1$.
4.
Quotient: If $l_2 \neq 0, \underset{P \rightarrow P_0}{\text{lim}}\frac {f(P)}{g(P)}=
\frac {l_1}{l_2}$.
5.
Powers: If $\alpha$ is a rational number and if l1 >0, then $\underset{P \rightarrow P_0}{\text{lim}} f(P)^{\alpha}= l_1^{\alpha}$.

Example 4.8       

Example 4.9   Compute the limit at (0,0) of the function f such that

\begin{displaymath}f(x,y)= \frac {xy-y^2}{\sqrt{x}-\sqrt{y}}
\end{displaymath}

Both the numerator and the denominator approach 0, when (x,y) approaches the origin, thus we cannot use the quotient rule of prop.  4.7. We multilply the numerator and the deniminator by the conjugate expression of the denominator; we have:

\begin{displaymath}f(x,y)= \frac {xy-y^2}{\sqrt{x}-\sqrt{y}}
= \frac {(xy-y^2)(...
...\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}
= y(\sqrt{x}+\sqrt{y})
\end{displaymath}

Hence:

\begin{displaymath}\underset{(x,y) \rightarrow (0,0)}{\text{lim}} f(x,y)= \under...
...y) \rightarrow (0,0)}{\text{lim}} ( y(\sqrt{x}+\sqrt{y}) ) =0.
\end{displaymath}

Proposition 4.10   If it exists, the limit of f at (x0,y0) is unique.

The main usage of that proposition is in disproving the existence of a limit.

Example 4.11   Let $f(x,y)=\frac {x^2-y^2}{x^2+y^2}$ define a function on the whole plane, but not at the origin.

We will approach the origin on a line, whose equation is y=mx. Then:

\begin{displaymath}f(x,y) = f(x,mx)= \frac {x^2-(mx)^2}{x^2+(mx)^2} = \frac {x^2(1-m^2)}{x^2(1+m^2)}
=\frac {1-m^2}{1+m^2}
\end{displaymath}

If the values m1 and m2 of the parameter m verify the condition $M_1 \neq \pm m_2$, the limits of f(x,mx) at 0 are diferent. Therefore, the function f has no limit at the origin.


next up previous contents
Next: Continuous functions. Up: Multivariable functions. Previous: Level curves.
Noah Dana-Picard
2001-05-30
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