1^ste manier : zonder de regel van de l'Hospital
\( \scriptsize \quad \scriptsize \displaystyle \lim_{x \to 1}\: \frac{ \sqrt{2\!+\!\sqrt{x+3}}-\!2}{x-1}\\ = \scriptsize \displaystyle \lim_{x \to 1}\: \frac{(\sqrt{2\!+\!\sqrt{x+3}}-\!2). (\sqrt{2\!+\!\sqrt{x+3}}+\!2)} {(x-1).(\sqrt{2\!+\!\sqrt{x+3}}+\!2)}\\ = \scriptsize \displaystyle \lim_{x \to 1}\: \frac {2+ \sqrt{x+3}-4}{(x-1).(\sqrt{2\!+\!\sqrt{x+3}}+\!2)}\\ = \scriptsize \displaystyle \lim_{x \to 1}\: \frac {\sqrt{x+3}-2}{(x-1).(\sqrt{2\!+\!\sqrt{x+3}}+\!2)}\\ = \scriptsize \displaystyle \lim_{x \to 1}\: \frac {x+3-4}{(x-1).(\sqrt{2\!+\!\sqrt{x+3}}+\!2).(\sqrt{x+3}+2)}\\ = \scriptsize \displaystyle \lim_{x \to 1}\: \frac {1}{(\sqrt{2\!+\!\sqrt{x+3}}+\!2).(\sqrt{x+3}+2)}\\ = \frac {1}{(2+2).(2+2)} = \frac {1}{16} \)
2^de manier : met de regel van de l'Hospital
\( \scriptsize \quad \scriptsize \displaystyle \lim_{x \to 1}\: \frac{\sqrt{2\!+\!\sqrt{x+3}}-\!2}{x-1}\left ( =\frac 00 \right )\overset{H}{=} \scriptsize \displaystyle \lim_{x \to 1}\: D\: \sqrt{2\!+\!\sqrt{x+3}}\\ = \scriptsize \displaystyle \lim_{x \to 1}\:\left (\frac{1}{2.\sqrt{2\!+\!\sqrt{x+3}}}.\frac{1}{2.\sqrt{x+3}}\right )=\frac{1}{2.\sqrt{2+2}}.\frac{1}{2.\sqrt 4}=\frac{1}{16} \)