<s>[INST] Not sure how to solve $\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$ 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.
 [/INST] 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 $a>0,$ $$f(a)=\dfrac1{2a}$$
$$\dfrac{f(1)}{f(4)}=?$$
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