L'hopital's rule mistake for multivariable calculus I've been helping some students in multivariable calculus and they're currently working with limits and it's been expressed by the lecturer that L'hopital's is exclusively for single-variable functions. I know this to be true as well, but the students have found that using L'hopital's in the same way that they take partial derivatives, leads them to solutions to their limit problems.
Take for example
$$\lim_{(x,y)\to(0,0)} \frac{e^{xy}-1}{y}$$
If we plug in our limit, we get an indeterminate form. Students have found that if they "fix $x$" and use L'hopital's in $y$, they get
$$\lim_{(x,y)\to(0,0)} \frac{xe^{xy}}{1} = 0\cdot 1 = 0$$
which is in fact the limit when you do it through less dubious means.
It's worked in other problems as well, even those where the limit doesn't exist, students have just used L'hopital's with respect to different variables and found that they "gave different limits in the end, so the overall limit must not exist", which for that problem, it definitely did not exist by checking a few simple paths.
Here are my questions.

*

*Why does this seem to work sometimes?

*How would you convince students that this is not the way to go? I've tried with counterexamples, but they seem to still prefer it because it works on all their actual homework problems (Marsden).

 A: Let us consider these two solutions of the problem above which could be asked in our Analysis III course's midterm exam: Find the limit $\lim_{(x,y)\rightarrow (0,0)}\frac{e^{xy}-1}{y}$, if exists. Suppose it exists and call it $L$.
$\begin{array}\\
\underline{\text{L'hospital rule method}}  \hspace{1 cm} &\text{V. S.}& \hspace{1 cm} \underline{\text{Series method}}\\
L=\lim_{(x,y)\rightarrow (0,0)}\frac{e^{xy}-1}{y} \hspace{1 cm}  &\text{V. S.}&  \hspace{1 cm}L=\lim_{(x,y)\rightarrow (0,0)}\frac{e^{xy}-1}{y}\\
\text{L'hospital rule w.r.t. y}\hspace{1 cm} &\text{V. S.}& \hspace{1 cm} \text{Maclaurin Series at y=0}\\
L=\lim_{(x,y)\rightarrow (0,0)}\frac{xe^{xy}}{1}  &\text{V. S.}& \ \hspace{1 cm}L=\lim_{(x,y)\rightarrow (0,0)}\frac{(1+xy+\frac{1}{2}x^2y^2+...)-1}{y}\\
L=\lim_{(x,y)\rightarrow (0,0)} xe^{xy}  &\text{V. S.}& \hspace{1 cm}L=\lim_{(x,y)\rightarrow (0,0)}\frac{xy+\frac{1}{2}x^2y^2+\frac{1}{6}x^3y^3...}{y}\\
L=0\times e^0  &\text{V. S.}& \hspace{1 cm} L=\lim_{(x,y)\rightarrow (0,0)}\frac{y(x+\frac{1}{2}x^2y+\frac{1}{6}x^3y^2...)}{y}\\
L=0\times 1  &\text{V. S.}& \hspace{1 cm}L=\lim_{(x,y)\rightarrow (0,0)}x+\frac{1}{2}x^2y+\frac{1}{6}x^3y^2...\\
L=0  &\text{V. S.}& \hspace{1 cm} L=0\\
\end{array}$
Now, let's consider the following problem: Find the limit $\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x^2+y^2+y}$, if exists. Suppose it exists and call it $L$.
$\begin{array}\\
\underline{\text{L'hospital rule method}}  \hspace{1 cm} &\text{V. S.}& \hspace{1 cm} \underline{\text{Series method}}\\
L=\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x^2+y^2+y} \hspace{1 cm}  &\text{V. S.}&  \hspace{1 cm}L=\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x^2+y^2+y}\\
\text{L'hospital rule w.r.t. y}\hspace{1 cm} &\text{V. S.}& \hspace{1 cm} \text{Maclaurin Series at y=0}\\
L=\lim_{(x,y)\rightarrow (0,0)}\frac{x}{2y+1} \hspace{1 cm} &\text{V. S.}& \hspace{1 cm}\text{They are polynomials. Good.:)}\\
L=\frac{0}{2\times 0+1} \hspace{1 cm} &\text{V. S.}& \hspace{1 cm}\text{There are no cancellations. What to do?:(}\\
L=0 \,\,\,\text{No! This is wrong.}\hspace{1 cm} &\text{V. S.}& \hspace{1 cm} L=???\\
\end{array}$
The limit in the second question does not exist. Series method turned out to be better. It didn't lead us to a wrong conclusion. Taylor's series (https://en.wikipedia.org/wiki/Taylor_series) at $x=0$ of functions of one variable $x$ is called Maclaurin's series (https://mathworld.wolfram.com/MaclaurinSeries.html). We used the Maclaurin series of the function $e^t$ in the first problem, then put $t=xy$. This is actually the Taylor series of the two-variable function $f(x,y)=e^{xy}$ at $(x,y)=(0,0).$
Why do they usually ask two-variable limit questions at $(x,y)=(0,0)$? Because of translations. The limit of the function $f(x,y)$ at $(x,y)=(a,b)$ is equal to the limit of the function $f(x+a,y+b)$ at $(x,y)=(0,0)$. That is why Maclaurin's series are more famous than Taylor's series although they are very special case of Taylor's.
Functions of two and more variables are a diffucult subject. There are more to be said about their Taylor's series. They need time to study. Therefore, it is not easy to answer OP's first question immediately like when there are cancellations in series method, two ways give the same result. Question 2 might lead to a war. Do what you like.
