Solutions To Mathematics Textbooks/Calculus (3rd) (0521867444)/Chapter 5

Question 1

i

$\underset{x\to 1}{\lim }\frac{x^2-1}{x+1}=\underset{x\to 1}{\lim }\frac{\cancel{(x+1)}(x-1)}{\cancel{x+1}}$

$=\underset{x\to 1}{\lim }x-1$

$=0$