If you cannot make any headway with \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\)
in rectangular coordinates, try changing to polar coordinates. Substitute \(x=r
\cos \theta, y=r \sin \theta,\) and investigate the limit of the resulting
expression as \(r \rightarrow 0 .\) In other words, try to decide whether there
exists a number \(L\) satisfying the following criterion: $$|r|<\delta
\Rightarrow|f(r, \theta)-L|<\epsilon$$ If such an \(L\) exists, then $$\lim
_{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{r \rightarrow 0} f(r \cos \theta, r
\sin \theta)=L$$ For instance, $$\lim _{(x, y) \rightarrow(0,0)}
\frac{x^{3}}{x^{2}+y^{2}}=\lim _{r \rightarrow 0} \frac{r^{3} \cos ^{3}
\theta}{r^{2}}=\lim _{r \rightarrow 0} r \cos ^{3} \theta=0$$ To verify the
last of these equalities, we need to show that Equation
(1) is satisfied with \(f(r, \theta)=r \cos ^{3} \theta\) and \(L=0 .\) That is,
we need to show that given any \(\epsilon>0,\) there exists a \(\delta>0\) such
that for all \(r\) and \(\theta\) $$|r|<\delta \Rightarrow\left|r \cos ^{3}
\theta-0\right|<\epsilon$$ since $$\left|r \cos ^{3}
\theta\right|=|r|\left|\cos ^{3} \theta\right| \leq|r| \cdot 1=|r|$$ the
implication holds for all \(r\) and \(\theta\) if we take \(\delta=\epsilon\) In
contrast, $$\frac{x^{2}}{x^{2}+y^{2}}=\frac{r^{2} \cos ^{2}
\theta}{r^{2}}=\cos ^{2} \theta$$takes on all values from 0 to 1 regardless of
how small \(|r|\) is, so that \(\lim _{(x, y) \rightarrow(0,0)} x^{2}
/\left(x^{2}+y^{2}\right)\) does not exist.
In each of these instances, the existence or nonexistence of the limit as \(r
\rightarrow 0\) is fairly clear. Shifting to polar coordinates does not always
help, however, and may even tempt us to false conclusions. For example, the
limit may exist along every straight line (or ray)
\(\theta=\) constant and yet fail to exist in the broader sense. Example 5
illustrates this point. In polar coordinates, \(f(x, y)=\left(2 x^{2} y\right)
/\left(x^{4}+y^{2}\right)\) becomes $$f(r \cos \theta, r \sin \theta)=\frac{r
\cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\sin ^{2} \theta}$$for \(r
\neq 0 .\) If we hold \(\theta\) constant and let \(r \rightarrow 0,\) the limit is
\(0 .\) On the path \(y=x^{2},\) however, we have \(r \sin \theta=r^{2} \cos ^{2}
\theta\) and $$\begin{aligned}
f(r \cos \theta, r \sin \theta) &=\frac{r \cos \theta \sin 2 \theta}{r^{2}
\cos ^{4} \theta+\left(r \cos ^{2} \theta\right)^{2}} \\
&=\frac{2 r \cos ^{2} \theta \sin \theta}{2 r^{2} \cos ^{4} \theta}=\frac{r
\sin \theta}{r^{2} \cos ^{2} \theta}=1
\end{aligned}$$ Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show
that the limit does not exist.
$$f(x, y)=\frac{2 x}{x^{2}+x+y^{2}}$$
Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine 
