Exercise:

(a) Explain why you cannot apply the Direct Substitution Property to evaluate the limit:

lim x5 x 2 6x + 5 x 5(b) What other method can you use to evaluate the above limit?

(c) What is wrong with the following equation?

x 2 6x + 5 = x - 1 x 5

(d) Explain why the equation

lim x5 x 2 6x + 5 = lim x5 (x 1) is correct. x 5 (e) Evaluate the limit:

lim x5 x 2 6x + 5 x 5

Exercises:

True or False: explain why it is True, or give a counterexample if it is False.

1. limx 4 ( 2 xx4 8x4 )=limx 4 2 xx4lim x 4 8x4

2. limx 1x2+6 x7x2+5 x6

=lim x1 ( x

2+6 x7)lim x1 ( x

2+5 x6)

3. limx 1x3

x2+2 x4=

limx 1 (x3)lim x1 ( x

2+2 x4)

4. If limx 5 f ( x)=2 and limx 5 g (x)=0, then limx 5 [ f (x)g (x) ] does not exist. 5. If limx 5 f (x)=0 and limx 5 g (x)=0, then limx 5 [ f (x)g (x) ] does not exist.6. If limx 6 [ f (x)g (x)] exists, then the limit must be f (6)g (6) .

7. If p is a polynomial, then limx a p( x)= p(a) .