Page 68 - Calculus Workbook For Dummies

52 Part II: Limits and Continuity

3. Cancel.

- 1
= lim
x " 2 2 x
4. Plug in.
- 1
=
4
1
= -
4
i lim x = –36
x " 0 1 + 1
6 x - 6
Multiply by the least common denominator, multiply out, cancel, plug in.

x 6^ x - 6h
= lim $
x " 0 1 + 1 6^ x - 6h
6 x - 6
x x -
6 ^ 6h
= lim
6 +
x " ^ x - h 6
0
x x -
6 ^ 6h
= lim x
x " 0
= lim6^ x - 6h
x " 0
6 0 -
= ^ 6h
= - 36
j lim sinx = 1
x " 0 x
No work required — except for the memorization, that is.
*k lim x = 1
3
x " 0 sin x 3
Did you get it? If not, try the following hint before you read the solution: This fraction sort of
resembles the one in problem 10. Still stuck? Okay, here you go:
1. Multiply numerator and denominator by 3.
You’ve got a 3x in the denominator, so you need 3x in the numerator as well (to make the frac-
tion look more like the one in problem 10).
x 3
= lim $
x " 0 sin x3 3
3 x
= lim
x " 0 3 sin x3
1 1
2. Pull the through the lim symbol (the 3 in the denominator is really a , right?).
3 3
1 3 x
= lim
3
3 x " 0 sin x
Now, if your calc teacher lets you, you can just stop here — since it’s “obvious” that
3 x 1 1
lim = 1 — and put down your final answer of $ 1, or . But if your teacher’s a stickler
3
x " 0 sin x 3 3
for showing work, you’ll have to do a couple more steps.
3. Set u 3= x.