This is a Japanese highschool homework assignment that has stumped everyone
$$ \lim_{h\to 0}{\frac{\log\left( x+h+\sqrt{(x+h)^2 + a} \right) - \log\left(x+\sqrt{x^2 + a}\right)}{h}} $$
I'm pretty rusty but from what I remember it's possible to move some terms around and I've gotten it to like
$$ \lim_{h\to 0}{\log\left(\left(\frac{x+h+\sqrt{(x+h)^2 + a}}{x+\sqrt{x^2 + a}}\right)^{1/h}\right)} $$
and it kind of reminds me of the formula
$$ \lim_{x\to 0}{(1+x)^{1/x}} $$
But I'm sure there's a twist in there somewhere. I would be grateful for any pointers and I'm not asking for a full solution but I'd be very grateful to know the trick?
Let $z=x+\sqrt{x^2+a}$, then $$L=\lim_{h \to 0}\frac{1}{h} [\ln(x+h+\sqrt{(x+h)^2+a})-\ln(x+\sqrt{x^2+a})]$$ $$L=\lim_{h\to 0}\frac{1}{h} \ln\frac{(x+h+\sqrt{x^2+a+2ahx+h^2}}{x+\sqrt{x^2+a}}$$ ignoring $h^2$ as we have only a linear term in $h$ and the binomial approximation $(1+y)^{\nu} \approx (1+\nu y)$, when $|y|$ is as small as we please, then $$L=\lim_{h \to 0} \frac{1}{h} \ln\left(\frac{(x+h+\sqrt{x^2+a}+hx/\sqrt{x^2+a}}{z}\right)$$ $$\implies L= \lim_{h \to 0}\frac{1}{h} \ln[1+h/z+hx/(z\sqrt{x^2+a})]$$ Using $\ln(1+y)\approx 1+y$, we get $$L=\lim_{h\to 0} \frac{1}{h} [h(1/z+x/(z\sqrt{x^2+a})]=\frac{1}{x+\sqrt{x^2+a}}\left(1+\frac{x}{\sqrt{x^2+a}}\right)$$ $$ \implies L=\frac{1}{\sqrt{x^2+a}}$$