I know this is an old question, but the accepted answer is simply wrong. A minus sign got lost somewhere. As \(x\to0\), the numerator approaches
\(\sqrt{0^2+16}-4 = \sqrt{16} - 4 = 4 - 4 = 0\)
while the denominator a non-zero (complex!) number, so the overall limit should be 0.
However, I suspect there may have been a typo in the original question, and it was intended to say
\(\displaystyle \lim_{x\to0} \frac{\sqrt{x^2+16} - 4}{\sqrt{x^2 + 4} - 2}\)
Now evaluating at \(x=0\), the limit has the indeterminate form 0/0 - a much more interesting result! To evaluate the limit, recall the difference of squares identity,
\(a^2 - b^2 = (a - b) (a + b)\)
Rewrite the limit as
\(\displaystyle \lim_{x\to0} \frac{a - b}{c - d} = \lim_{x\to0} \frac{a^2-b^2}{c^2-d^2} \cdot \frac{c+d}{a+b}\)
where
\(\begin{cases} a = \sqrt{x^2+16} \\ b = 4 \\ c=\sqrt{x^2+4} \\ d = 2\end{cases}\)
Then
\(\dfrac{a^2 - b^2}{c^2 - d^2} \cdot \dfrac{c+d}{a+b} = \dfrac{(x^2+16)-16}{(x^2-4)-4} \cdot \dfrac{\sqrt{x^2+4}+2}{\sqrt{x^2+16}+4} = \dfrac{\sqrt{x^2+4}+2}{\sqrt{x^2+16}+4}\)
By canceling the factors of \(x^2\), we've removed the discontinuity at \(x=0\). So
\(\displaystyle \lim_{x\to0} \frac{\sqrt{x^2+16} - 4}{\sqrt{x^2 + 4} - 2} = \lim_{x\to0} \frac{\sqrt{x^2+4} + 2}{\sqrt{x^2+16}+4} = \frac{\sqrt{0^2+4}+2}{\sqrt{0^2+16}+4} = \frac48 = \boxed{\frac12}\)
