Not sure how to solve $\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$

Question

So I got this problem:

Determine the following limit value: $$\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$$

What I tried is:

$\large{\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}\cdot{\sqrt{x^2+16}+4\over\sqrt{x^2+16}+4}=\\\lim_\limits{x\to0}{({\sqrt{x^2+1}-1})(\sqrt{x^2+16}+4)\over x^2+16-16}}$

From this point I just get everything messy, and can't get anything out of it, so I believe this is not the way to solve this. Doing it from a table of values gives 4, but I should solve this without the table.

Answer 1

Now, you can make the following: $$\frac{(\sqrt{x^2+1}-1)(\sqrt{x^2+16}+4)}{x^2+16-16}=\frac{(x^2+1-1)(\sqrt{x^2+16}+4)}{x^2(\sqrt{x^2+1}+1)}\rightarrow\frac{4+4}{1+1}=4.$$

Answer 2

Hint:

$$f(a)=\lim_{x\to0}\dfrac{\sqrt{x^2+a^2}-a}{x^2}=\lim...\dfrac{x^2+a^2-a^2}{x^2(\sqrt{x^2+a^2}+a)}=\dfrac1{\sqrt{a^2}+a}$$

For $a>0,$ $$f(a)=\dfrac1{2a}$$

$$\dfrac{f(1)}{f(4)}=?$$