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

= Question 1 =

i
$$\lim_{x\to1}{\frac{x^2-1}{x+1}} = \lim_{x\to1}{\frac{\cancel{(x+1)}(x-1)}\cancel{x+1}} $$

$$= \lim_{x\to1} {x-1}$$

$$= 0$$