A limit involving sinh I'm trying to show that $$\lim_{u\to 0}\frac{\partial}{\partial u}\frac{\sinh(y\sqrt{2u}) + \sinh(x\sqrt{2u})}{\sinh((x + y)\sqrt{2u})} = -xy$$
The method I was trying resulted in pages and pages of messy computations, and I'm doubtful that this is the best way to go about it. Any ideas would be appreciated. 
-- Thanks. 
 A: Method Guide


*

*Expand the numerator and denominator into 2-term Taylor expansions.

*Justify that the remaining terms of the numerator and denominator Taylor expansions can be discarded because the limit as $u \to 0$ kills them. So it suffices to use these expansions.

*You should get
$$ \frac{(x+y) + \frac{(x^3 + y^3)u}{3}}{(x+y) + \frac{(x+y)^3u}{3}}$$
after you cancel and simplify a bit.

*Notice that for general $f(u) = \frac{a+bu}{c+du}$, we get that $f'(u) = \frac{bc-ad}{(c+du)^2}$, so that as $u \to 0$, we get $\frac{bc-ad}{c^2}$

*Plug in our situation above to get the result you want.


I carried this through, and I was a bit surprised to see if fall out so nicely at the end.
Addendum on the Taylor Series of $\sinh x$
Recall the Taylor series for $e^x$:
$$
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots
$$
As $\sinh x = \dfrac{e^x - e^{-x}}{2},$ we use the Taylor series for $e^x$ to see that
$$
\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots
$$
This allows you to expand the numerator and denominator into Taylor polynomials. 
A: Let $a=(x-y)/\sqrt2$ and $b=(x+y)/\sqrt2$.  Then
\begin{align*}
&\lim_{u\to0}\frac{\partial}{\partial u} \frac{\sinh(x\sqrt{2u}) + \sinh(y\sqrt{2u})}{\sinh((x+y)\sqrt{2u})} \\
&= \lim_{u\to0}\frac{\partial}{\partial u} \frac{\cosh(a\sqrt u)}{\cosh(b\sqrt u)} \\
&= \lim_{u\to0}\frac{a\Bigl(\overbrace{\frac{\sinh(a\sqrt u)}{2\sqrt u}}^{\to a/2}\Bigr)\cosh(b\sqrt u) - b\cosh(a\sqrt u)\Bigl(\overbrace{\frac{\sinh(b\sqrt u)}{2\sqrt u}}^{\to b/2}\Bigr)}{\cosh^2(b\sqrt u)} \\
&= \frac{a^2-b^2}{2} \\
&= -xy
\end{align*}
