Prova Cálculo 1 A Gabarito - Prof. Sacha Friedli 2015/1

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❈á❧❝✉❧♦ ■✿ ✶❛ ♣r♦✈❛✱ ✶✶✴✵✹✴✷✵✶✺✱ ✽❤✵✵✱ ❉✉r❛çã♦✿ ✶❤✹✵✳ ❆
✶✳ (6♣ts) ❘❡s♦❧✈❛ ❛ s❡❣✉✐♥t❡ ❞❡s✐❣✉❛❧❞❛❞❡✿ 1x+1 <
1
x+2 ✳
P❛ss❛♥❞♦ ♣♦r ❡①❡♠♣❧♦ ♦
1
x+2 ❞♦ ❧❛❞♦ ❡sq✉❡r❞♦ ❡ ❝♦❧♦❝❛♥❞♦ ♥♦ ♠❡s♠♦ ❞❡♥♦♠✐♥❛❞♦r✱ ❛ ❞❡s✐❣✉❛❧❞❛❞❡ s❡ t♦r♥❛
1
(x+1)(x+2) < 0
(3♣ts)✳ ▼♦♥t❛♥❞♦ ❛ t❛❜❡❧❛ ❞❡ s✐♥❛✐s ❞♦s t❡r♠♦s x+ 1 ❡ x+ 2 ♦❜t❡♠♦s S = (−2,−1) (3♣ts)✳ ✭▲❡♠❜r♦ q✉❡ é ♠❡❧❤♦r ❡✈✐t❛r
♠✉❧t✐♣❧✐❝❛r ❛♠❜♦s ❧❛❞♦s ❞❡ ✉♠❛ ❞❡s✐❣✉❛❧❞❛❞❡ ♣♦r ❛❧❣♦ q✉❡ ❞❡♣❡♥❞❡ ❞❡ x✱ ♣♦✐s ♣r❡❝✐s❛ s❛❜❡r s❡ ♠✉❞❛ ♦✉ ♥ã♦ ♦ s✐♥❛❧ ❞❛
❞❡s✐❣✉❛❧❞❛❞❡✳✮
✷✳ (6♣ts) ▼♦♥t❡ ❡ ❝❛❧❝✉❧❡ ✉♠ ❧✐♠✐t❡ q✉❡ r❡♣r❡s❡♥t❡ ❛ ✐♥❝❧✐♥❛çã♦ ❞❛ r❡t❛ t❛♥❣❡♥t❡ ❛♦ ❣rá✜❝♦ ❞❛ ❢✉♥çã♦ f(x) =
√
1− x ♥♦ ♣♦♥t♦
P = (0, 1)✳
❖ ❧✐♠✐t❡ ♣r♦❝✉r❛❞♦ é
lim
λ→0
f(λ)− f(0)
λ− 0 = limλ→0
√
1− λ−√1
λ
, (3♣ts)
❡ é ❞❛ ❢♦r♠❛ ✏0/0✑✳ ▼✉❧t✐♣❧✐❝❛♥❞♦ ❡ ❞✐✈✐❞✐♥❞♦ ♣❡❧♦ ❝♦♥❥✉❣❛❞♦ ♦❜t❡♠♦s
lim
λ→0
√
1− λ− 1
λ
= lim
λ→0
√
1− λ− 1
λ
√
1− λ+ 1√
1− λ+ 1(1♣ts) = limλ→0
−1√
1− λ+ 1(1♣ts) = −
1
2 , (1♣ts)
✸✳ (8♣ts) ❯s❛♥❞♦ s♦♠❡♥t❡ ♠ét♦❞♦s q✉❡ ❢♦r❛♠ ✈✐st♦s ❛té ❛❣♦r❛✱ ❝❛❧❝✉❧❡✿ limx→0
(senx)2
x tan x ✱ limx→−4+
x2−4x−5
x2+5x+4 ✳
❖ ♣r✐♠❡✐r♦ ❧✐♠✐t❡ é ❞❛ ❢♦r♠❛ ✏
0
0 ✑✳ ▼❛s ♦❜s❡r✈❡ q✉❡
(senx)2
x tanx
=
senx
x
· cosx .
▲♦❣♦✱ ♣❡❧❛ ♣r♦♣r✐❡❞❛❞❡ ❞♦ ❧✐♠✐t❡✱
lim
x→0
(senx)2
x tanx
=
{
lim
x→0
senx
x
}
· lim
x→0
cosx(2♣ts) = 1 · 1 = 1 .(2♣ts)
◆♦ s❡❣✉♥❞♦ ❧✐♠✐t❡ t❡♠♦s limx→−4+{x2 − 4x− 5} = 27 6= 0✱ limx→−4+{x2 + 5x+ 4} = 0✱ ❡♥tã♦ ♦ ❧✐♠✐t❡ ❞❡✈❡ s❡r ♦✉ +∞✱ ♦✉
−∞✳ P❛r❛ ❞❡t❡r♠✐♥❛r q✉❛❧✱ ♣♦❞❡♠♦s ❢❛t♦r❛r ♦ ❞❡♥♦♠✐♥❛❞♦r ♣❛r❛ ❡♥t❡♥❞❡r ♠❡❧❤♦r ♦ s❡✉ s✐♥❛❧✱ ❡ ❡s❝r❡✈❡r
x2 − 4x− 5
x2 + 5x+ 4
=
x2 − 4x− 5
x+ 1
· 1
x+ 4
.
❆❣♦r❛✱ ❝♦♠♦ limx→−4+
x2−4x−5
x+1 = − 273 6= 0 ❡ limx→−4+ 1x+4 = +∞✱ t❡♠♦s ✭✈❡❥❛ ❛ Pr♦♣♦s✐çã♦ ✹✳✷✮
lim
x→−4+
x2 − 4x− 5
x+ 1
1
x+ 4
= −∞(4♣ts)
✹✳ (5♣ts) ❆ ❢✉♥çã♦ f é ❝♦♥tí♥✉❛ ❡♠ x = 2❄ ❏✉st✐✜q✉❡✳
f(x) =
{
|x−2|
2x−4 s❡ x 6= 2 ,
1
2 s❡ x = 2 .
Pr❡❝✐s❛♠♦s ✈❡r✐✜❝❛r s❡ limx→2 f(x) ❡①✐st❡ ❡ é ✐❣✉❛❧ ❛ f(2)✳ ❚❡♠♦s f(2) =
1
2 ♣♦r ❞❡✜♥✐çã♦✱ ♠❛s ♦❜s❡r✈❡ q✉❡
lim
x→2+
f(x) = lim
x→2+
|x− 2|
2x− 4 = limx→2+
✘
✘✘x− 2
2
✘
✘
✘✘(x− 2) =
1
2 , (2♣ts)
❡ q✉❡
lim
x→2−
f(x) = lim
x→2−
|x− 2|
2x− 4 = limx→2−
−
✘
✘
✘✘(x− 2)
2
✘
✘
✘✘(x− 2) = −
1
2 .(2♣ts)
▲♦❣♦✱ ♦ ❧✐♠✐t❡ ❜✐❧❛t❡r❛❧ limx→2 f(x) ♥❡♠ ❡①✐st❡✱ ❡ ♣♦rt❛♥t♦ f ♥ã♦ é ❝♦♥tí♥✉❛ ❡♠ x = 2 (1♣ts)✳ ✭❖❜s❡r✈❡✱ ♥♦ ❡♥t❛♥t♦✱ q✉❡ f
é ❝♦♥tí♥✉❛ ❛ ❞✐r❡✐t❛✳✮
✺✳ ✭❇❖◆❯❙✮ ✭♥♦ ♠á①✐♠♦ (5♣ts)✮ ❉ê ❛ ❞❡✜♥✐çã♦ r✐❣♦r♦s❛ ❞♦ s✐❣♥✐✜❝❛❞♦ ❞❡ ✏f(x) t❡♥❞❡r ❛ L q✉❛♥❞♦ x t❡♥❞❡ ❛ +∞✑✳ ❊♠
s❡❣✉✐❞❛✱ ✉s❛♥❞♦ ❛ ❞❡✜♥✐çã♦✱ ♠♦str❡ q✉❡ f(x) = 12x3 t❡♥❞❡ ❛ 0 q✉❛♥❞♦ x t❡♥❞❡ ❛ +∞✳
❉✐③ s❡ q✉❡ ✏f(x) t❡♥❞❡ ❛ L q✉❛♥❞♦ x t❡♥❞❡ ❛ +∞✑ s❡ ♣❛r❛ t♦❞♦ ǫ > 0 ❡①✐st❡ ✉♠ ♥ú♠❡r♦ N t❛❧ q✉❡ f(x)− L ≤ ǫ ♣❛r❛ t♦❞♦
x ≥ N ✳ ❊s❝r❡✈❡✲s❡✿ limx→∞ f(x) = L✳ ◆♦ ❝❛s♦✱ ✜①❡♠♦s ✉♠ ǫ > 0 q✉❛❧q✉❡r ❡ ♣r♦❝✉r❡♠♦s s❛❜❡r s❡ |f(x) − 0| = 12|x3| ≤ ǫ
♣❛r❛ t♦❞♦ x s✉✜❝✐❡♥t❡♠❡♥t❡ ❣r❛♥❞❡✳ ❈♦♠♦ ❛q✉✐❧♦ é ✈❡r❞❛❞❡ s❡ ❡ s♦♠❡♥t❡ s❡ x ≥ (2ǫ)−1/3✱ ♣♦❞❡♠♦s s✐♠♣❧❡s♠❡♥t❡ ♣❡❣❛r
N = (2ǫ)−1/3✳