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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 6
}
\bigskip
{\bf 1.} 
Find the limit if it exists, or show that the
limit does not exist.
$$
\lim_{(x,y) \rightarrow (0,0)} {{2x} \over {2x+3y}}
\quad .
$$

{\bf Sol.} We first try to {\it plug-it-in}, and get $0/0$,
which is indeterminate.

 We next try to prove that the limit does {\bf not} exist
by exploring the limits when approaching $(0,0)$ along the
line $y=mx$.
$$
\lim_{ {{(x,y) \rightarrow (0,0)} \atop {y=mx}}} {{2x} \over {2x+3y}}
=
\lim_{x \rightarrow 0} {{2x} \over {2x+3mx}}=
$$
$$
\lim_{x \rightarrow 0} {{2x} \over {(2+3m)x}}=
\lim_{x \rightarrow 0} {{2} \over {(2+3m)}}=
{{2} \over {(2+3m)}} \quad .
$$
Since this {\bf depends} on $m$, the limit {\bf does not exist},
since different paths to $(0,0)$ yield different limits, and
there is no consensus.

{\bf Common mistakes}: {\bf 1.} 
$\lim_{x \rightarrow 0} {{2} \over {(2+3m)}}$.
This is correct, but not the final answer. The final answer should
not have $\lim$ in it!

{\bf 2.}
Find the limit if it exists, or show that the
limit does not exist.
$$
\lim_{(x,y) \rightarrow (0,0)} {{x^5} \over {x^2+y^2}} \quad .
$$

{\bf Solution}: Once again it gives $0/0$. If we do as above, we would
get that the limit is $0$, no matter on which $y=mx$ you are travelling to $(0,0)$.
So the limit {\bf probably} exists, and if it does it is equal to $0$.
But this is {\bf not} a conclusive proof. To prove it conclusively,
we convert to {\bf polar} coordinates:
$$
x=r \cos \theta \quad , \quad  y=r \cos \theta \quad .
$$
Recall that $x^2+y^2=r^2$, so
$$
\lim_{(x,y) \rightarrow (0,0)} {{x^5} \over {x^2+y^2}} 
=\lim_{r \rightarrow 0} {{(r \cos \theta)^5} \over {r^2}}=
=\lim_{r \rightarrow 0} {{r^5 \cos^5 \theta} \over {r^2}}=
$$
$$
\lim_{r \rightarrow 0} r^3 \cos^5 \theta
=
\cos^5 \theta \lim_{r \rightarrow 0} r^3 =
\cos^5 \theta \cdot 0 = 0 \quad .
$$

and (if the limit is with $(x,y)$ it can never have $x$ or $y$!).
\end





