Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (b)/Solution 1

If you try to evaluate the limit directly, you end up with 0/0, which means we cannot conclude anything for the moment, more work needs to be done. Since we have a square root, we will multiply on each side of the fraction by the conjugate in the hope to simplify things:

${\begin{aligned}\lim _{x\to 1}{\frac {{\sqrt {x+3}}-2}{x-1}}&=\lim _{x\to 1}{\frac {({\sqrt {x+3}}-2)}{(x-1)}}{\frac {({\sqrt {x+3}}+2)}{({\sqrt {x+3}}+2)}}\\&=\lim _{x\to 1}{\frac {(x+3)-4}{(x-1)({\sqrt {x+3}}+2)}}\\&=\lim _{x\to 1}{\frac {x-1}{(x-1)({\sqrt {x+3}}+2)}}\\&=\lim _{x\to 1}{\frac {1}{({\sqrt {x+3}}+2)}}\\&={\frac {1}{4}}\end{aligned}}$