Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (a)/Solution 1

Directly plugging in x=1 into the limit gives 0/0. We try to factor to cancel out the terms making the fraction zero.

$\begin{array}{rl}& \underset{x\to 1}{\lim }\frac{x^2+2x-3}{x-1}\\ =& \underset{x\to 1}{\lim }\frac{(x-1)(x+3)}{x-1}\\ =& \underset{x\to 1}{\lim }(x+3)\\ =& (1)+3\\ =& 4\end{array}$