Page 256 - Calculus Workbook For Dummies

240 Part IV: Integration and Infinite Series

q lim cosx π = - 1
x " / π 2 x -
2
0
1. Plug in: — onward!
0
2. Replace numerator and denominator with their derivatives: lim - sinx .
=
1
π x " / π 2
- sin
3. Plug in again: = 2 = - 1.
1
r lim 1 - cosx = 1
x " 0 x 2 2
0
1. Plug in: ; no worries.
0
2. Replace with derivatives: lim sinx .
=
x " 0 2 x
0
3. Plug in: again, so repeat.
0
4. Replace with derivatives again: lim cosx .
=
x " 0 2
1
5. Finish: = .
2
s lim ^` tanx - 1h sec x = 1
6 j
x " / π 4 3
$
1. Plugging in gives you 0 3, so on to Step 2.
tanx - 1 0
.
2. Rewrite: lim = : copasetic.
=
6
x " / π 4 cos x 0
2
sec x
3. Replace with derivatives: lim .
=
6
x " / π 4 - 6 sin x
π
sec 2
4. Plug in to finish: = 4 3 π = 2 = 1 .
3
6
- 6 sin
2
t lim c 1 + 1 m = - 3
x
x " 0 + cosx - 1
1. Plugging in gives you 3 - 3; no good.
1
cosx - + x 0
2. Rewrite by adding the fractions: lim . That’s a good bingo: .
=
x " 0 + x ^ cosx - 1h 0
- sinx + 1
3. Replace with derivatives: lim .
=
1 -
x " ^ cosx - h x sinx
0
+
1
4. Plug in to finish: = = - 3.
" 0 "
-
This 0 is “negative” because the denominator in the line just above is negative when x is
approaching zero from the right. By the way, don’t use “–0” in class — your teacher will call
a technical on you.
u lim cscx - logx = 3
_
i
x " 0 +
1. This limit equals 3 - - 3h, which equals 3 + 3 = 3.
^
2. You’re done! L’Hôpital’s Rule isn’t needed. You gotta be on your toes.
*v lim 1 + h 1 /x = e
x
^
x " 0
3
1. This is a 1 case — time for a new technique.
2. Set your limit equal to y and take the natural log of both sides.
1 /x
x
y = lim 1 + h
^
x " 0
1 /x
x
lny = ln lim 1 + h l
^
b
x " 0