Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (a)/Solution 2

Directly plugging in x=1 into the limit gives 0/0. Since this is an indeterminate form, we can use L'Hospital's rule:

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