# Evaluating $\lim_{h\to 0}\frac1h\left(\log(x+h+\sqrt{(x+h)^2+a})-\log(x+\sqrt{x^2+a})\right)$

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?

• from inspection, it seems like a substitution with a change of variable would help but I'm not sure how exactly Jun 19 '20 at 9:58
• It's just $f'(x)$ for $f(x)=\ln(x+\sqrt{x^2+a})$. Jun 19 '20 at 10:15
• actually looking at the original question it is just the definition of the derivative of $\log(x+\sqrt{x^2+a})$. sorry for a wrong suggestion Jun 19 '20 at 10:20
• Y'all are light years ahead of me.. if I'm trying to solve it from first principles, what would I be trying to achieve? Jun 19 '20 at 10:24
• Expression under limit can be written as $(1/h)\log(A/B)=(1/h)\log(1+((A-B)/B))$ and now use the limit $\lim_{t\to 0}\dfrac{\log(1+t)}{t}=1$ to reduce the expression to $\dfrac{A-B} {Bh}$ and proceed. Jun 19 '20 at 10:31

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 $$L=\lim_{h\to 0} \frac{1}{h} [h(1/z+x/(z\sqrt{x^2+a})]=\frac{1}{x+\sqrt{x^2+a}}\left(1+\frac{x}{\sqrt{x^2+a}}\right)$$ $$\implies L=\frac{1}{\sqrt{x^2+a}}$$