Lời giải:
Ta có:
\(\lim_{x\to 2}\frac{\sqrt{x^2+5}-3}{2-x}=\lim_{x\to 2}\frac{x^2+5-9}{(\sqrt{x^2+5}+3)(2-x)}=\lim_{x\to 2}.\frac{x^2-4}{(\sqrt{x^2+5}+3)(2-x)}\)
\(=\lim_{x\to 2}\frac{-(x+2)}{\sqrt{x^2+5}+3}=\frac{-(2+2)}{\sqrt{2^2+5}+3}=-\frac{2}{3}\)