Find the multivariable limit with exponential: $f(x,y) = \frac{e^{xy^2}-1}{x^2+y^2}$ at $(0,0)$ Let $f(x,y): \mathbb{R^2} \to \mathbb{R}$ where $f(x,y) = \frac{e^{xy^2}-1}{x^2+y^2}$.  If it exists, find $$\lim_{(x,y) \to (0,0)}\frac{e^{xy^2}-1}{x^2+y^2} $$I'm not entirely sure how to approach this.  I've tried using squeeze lemma to no avail.  (I could potentially get one of the variables out of the bottom, but that still doesn't help me.)  I also tried considering $(\alpha x,\beta y)$, but again, no success.  As well, i can't find a useful $u$ for a $u$ substitution.
Edit:
I'm trying to use power series expansion for $e^{xy^2}$.  I get: $$\lim_{(x,y) \to (0,0)}\frac{e^{xy^2}-1}{x^2+y^2}=\lim_{(x,y) \to (0,0)} \frac{1+xy^2+o(xy^2)-1}{x^2+y^2}\le \lim_{(x,y) \to (0,0)} \frac{xy^2+o(xy^2)}{y^2}=$$$$\lim_{(x,y) \to (0,0)} x+\frac{o(xy^2)}{y^2}$$  Now, i'm not sure how to deal with the $\frac{o(xy^2)}{y^2}$.  I'm not sure if i'm just digging myself a deeper hole, or if i'm getting somewhere with this technique.
 A: I think that
your start is reasonable.
To get a bound on
the error in the
power series for $e^x$
after $n$ terms,
$\begin{array}\\
e^x
&=\sum_{k=0}^{\infty} \frac{x^k}{k!}\\
&=\sum_{k=0}^{n-1} \frac{x^k}{k!}+\sum_{k=n}^{\infty} \frac{x^k}{k!}\\
\end{array}
$
so,
if $|x| < 1$,
$\begin{array}\\
|e^x-\sum_{k=0}^{n-1} \frac{x^k}{k!}|
&=|\sum_{k=n}^{\infty} \frac{x^k}{k!}|\\
&\le|\frac1{n!}\sum_{k=n}^{\infty} x^k|\\
&\le|\frac{x^n}{(1-x)n!}|\\
\end{array}
$
Therefore,
if $|xy^2| < \frac12$,
$|e^{xy^2}-\sum_{k=0}^{n-1} \frac{(xy^2)^k}{k!}|
\le|\frac{(xy^2)^n}{(1-xy^2)n!}|
\le|\frac{2(xy^2)^n}{n!}|
$.
Setting $n=1$,
$|e^{xy^2}-1|
\le|2xy^2|
$
or
$\frac{|e^{xy^2}-1|}{x^2+y^2}
\le\frac{|2xy^2|}{x^2+y^2}
\le\frac{|2x|}{(x/y)^2+1}
\le|2x|
$
so the limit is zero.
A: Found an "easier" way to go about finding the limit of this function.  $$0 \le \lim_{(x,y)\to (0,0)} \frac{|e^{xy^2}-1|}{x^2+y^2} = \lim_{(x,y)\to (0,0)} \frac{|e^{xy^2}-1|}{|xy^2|}*\frac{|xy^2|}{x^2+y^2}$$
$$ =\lim_{(x,y)\to (0,0)} \frac{|e^{xy^2}-1|}{|xy^2|}*\lim_{(x,y)\to (0,0)} \frac{|xy^2|}{x^2+y^2}=\lim_{u \to 0}\frac{|e^u-1|}{u}*\lim_{(x,y)\to (0,0)} \frac{|xy^2|}{x^2+y^2}$$
$$=1*\lim_{(x,y)\to (0,0)} \frac{|xy^2|}{x^2+y^2}\le \lim_{(x,y)\to (0,0)} \frac{|x|y^2}{y^2}=\lim_{(x,y)\to (0,0)} |x| = 0$$
So, by the squeeze theorem, $f(x,y)$ $\to$ 0 as $(x,y) \to (0,0)$.
