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?

  • 1
    $\begingroup$ from inspection, it seems like a substitution with a change of variable would help but I'm not sure how exactly $\endgroup$
    – Mathsisfun
    Jun 19 '20 at 9:58
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    $\begingroup$ It's just $f'(x)$ for $f(x)=\ln(x+\sqrt{x^2+a})$. $\endgroup$ Jun 19 '20 at 10:15
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    $\begingroup$ 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 $\endgroup$
    – Mathsisfun
    Jun 19 '20 at 10:20
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    $\begingroup$ 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? $\endgroup$ Jun 19 '20 at 10:24
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    $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    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 $$L=\lim_{h\to 0} \frac{1}{h} [h(1/z+x/(z\sqrt{x^2+a})]=\frac{1}{x+\sqrt{x^2+a}}\left(1+\frac{x}{\sqrt{x^2+a}}\right)$$ $$ \implies L=\frac{1}{\sqrt{x^2+a}}$$

  • $\begingroup$ Wow this is on some next level -- $\endgroup$ Jun 19 '20 at 12:07
  • $\begingroup$ OK, I must give you kudos for this is absolutely the right answer though the process of arriving at it is unbelievable it simply boggles my mind and relies on two techniques I've never even heard of but it's elegant in its simplicity.. simply amazing. $\endgroup$ Jun 20 '20 at 6:22
  • $\begingroup$ @Alexandros K Thanks, but neither upvote or acceptance of solution yet ! $\endgroup$
    – Z Ahmed
    Jun 20 '20 at 8:48
  • $\begingroup$ You know how it goes -- first of all, I'm trying to absorb the solution, but secondly, I need to solve it in such a way as to be able to use only highschool mathematics.. $\endgroup$ Jun 20 '20 at 11:31
  • $\begingroup$ @Alexandros K Take x=4 ans a=9 the limit will be simply one 1/5. $\endgroup$
    – Z Ahmed
    Jun 20 '20 at 13:48

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