1 Prova C lculo I UFMG C12013 1 gabarito 1

✶✸ ❞❡ ❛❜r✐❧ ❞❡ ✷✵✶✸✱ ✽❤✵✵✱ ❉✉r❛çã♦✿ ✶❤✹✵✳

❈á❧❝✉❧♦ ■✿ ●❛❜❛r✐t♦ ✶❛ ♣r♦✈❛✱ ❚✉r♠❛s ❘✶✱ ❘✷✱ ❖▲✳

✶✳ ✭✼♣ts✮ ❘❡s♦❧✈❛

1 x ≥

1
2−x ✳

P❛r❛ ❝♦♠❡ç❛r✱ ♦❜s❡r✈❡ q✉❡ é ♣r❡❝✐s♦ ❝♦♥s✐❞❡r❛r s♦♠❡♥t❡
♠❡s♠♦ ❞❡♥♦♠✐♥❛❞♦r✱ ♦❜t❡♠♦s

2(x−1) x(x−2) ≥ 0 (1♣ts)✳

x = 0✱ x = 2✳

❈♦❧♦❝❛♥❞♦ ♦s t❡r♠♦s ❞♦ ♠❡s♠♦ ❧❛❞♦ ❡ ❝♦❧♦❝❛♥❞♦ ♥♦

▼♦♥t❛♥❞♦ ✉♠❛ ❧✐♥❤❛ ♣❛r❛ ❝❛❞❛ t❡r♠♦ q✉❡ ❞❡♣❡♥❞❡ ❞❡

0

1

x

−

x−1

−

−

x−2

−

−

2(x−1) x(x−2) −

+

2

+

0

0

+

+

+

+

−
−

0

+

❖❜s❡r✈❛çã♦✿ ▼✉❧t✐♣❧✐❝❛r ♦s ❞♦✐s ❧❛❞♦s ❞❛ ❞❡s✐❣✉❛❧❞❛❞❡ ♣♦r

♥ã♦ é ✈❡r❞❛❞❡ q✉❡

2

x(2 − x) ≥ 0

♣❛r❛ t♦❞♦

x✳

❛ ❞❡s✐❣✉❛❧❞❛❞❡

◗✉❡♠ ❢❡③ ✐ss♦ ❞❡✈❡ t❡r ❡♥❝♦♥tr❛❞♦ ❛ s♦❧✉çã♦ ❡rr❛❞❛

2 − x ≥ x é ✉♠ ❡rr♦✱ ♣♦✐s
(−∞, 1]✳ ◆❡st❡ ❝❛s♦ ❞❡♠♦s

♣♦♥t♦s✳

f (x) =

✷✳ ✭✽♣ts✮ ❉❡t❡r♠✐♥❡ t♦❞❛s ❛s ❛ssí♥t♦t❛s ❞❡

x2 −7x+10 x2 −2x ✳ ❆ r❡t❛

x=2

é ❛ssí♥t♦t❛ ✈❡rt✐❝❛❧❄ ❏✉st✐✜q✉❡✳

Pr♦❝✉r❡♠♦s ♣r✐♠❡✐r♦ ❛ssí♥t♦t❛s ❤♦r✐③♦♥t❛✐s✱ ❝❛❧❝✉❧❛♥❞♦ ❧✐♠✐t❡s q✉❛♥❞♦

limx→∞ f (x)✳ ◗✉❛♥❞♦ x t♦♠❛ ✈❛❧♦r❡s ❣r❛♥❞❡s✱ ♦s
❣r❛✉ 2 q✉❡ ❞❡✈❡♠ s❡r ♠❛✐s ✐♠♣♦rt❛♥t❡s✳ P♦rt❛♥t♦✱

x

t❡♥❞❡ ❛

+∞

❈♦♠♦ ❝❛❞❛ ✉♠ ❞♦s t❡r♠♦s

7/x✱ 10/x2

❡

2/x

−∞✳

P❛r❛ ❝♦♠❡ç❛r✱ ❝❛❧❝✉❧❡♠♦s

❝♦❧♦q✉❡♠♦s ❡❧❡s ❡♠ ❡✈✐❞ê♥❝✐❛✱ ❡ s✐♠♣❧✐✜q✉❡♠♦s✿

x2 (1 − x7 + x102 )
1 − x7 +
=
lim x→∞ x→∞ x2 (1 − x2 )
1 − x2

x→∞

❡

t❡r♠♦s ❞❡ ❣r❛✉ ♠❛✐♦r sã♦ ♠❛✐s ✐♠♣♦rt❛♥t❡s✳ ◆♦ ❝❛s♦✱ sã♦ ♦s t❡r♠♦s ❞❡

lim f (x) = lim

1✳

+

0

S = (0, 1] ∪ (2, ∞) (6♣ts)✳ x(2 − x) ♣❛r❛ ♦❜t❡r

P♦rt❛♥t♦✱ ♦ ❝♦♥❥✉♥t♦ ❞❡ s♦❧✉çõ❡s ❞❛ ❞❡s✐❣✉❛❧❞❛❞❡ é

x✱

t❡♥❞❡ ❛ ③❡r♦ q✉❛♥❞♦

x → ∞✱

10 x2 .

♦ ♥✉♠❡r❛❞♦r ❡ ♦ ❞❡♥♦♠✐♥❛❞♦r ❛♠❜♦s t❡♥❞❡♠ ❛

▲♦❣♦✱

lim f (x) =

x→∞
P♦rt❛♥t♦✱

❛ r❡t❛ ❤♦r✐③♦♥t❛❧ ❞❡ ❡q✉❛çã♦

t❛♠❜é♠ ♦s t❡r♠♦s ❞❡ ❣r❛✉

2

1
= 1 .(2.5♣ts)
1

y = 1 é ❛ssí♥t♦t❛ ❤♦r✐③♦♥t❛❧ ✭❛ ❞✐r❡✐t❛✮ (0.5♣ts)✳

lim f (x) =

x→−∞

x

t❡♥❞❡ ❛

−∞✱

sã♦

1
= 1.
1

P♦rt❛♥t♦✱

❛ r❡t❛ ❤♦r✐③♦♥t❛❧ ❞❡ ❡q✉❛çã♦

✈❡rt✐❝❛✐s✳

P❛r❛ ✐ss♦✱ ♦s ❝❛♥❞✐❞❛t♦s só ♣♦❞❡♠ ✈✐r ❞❡ ✉♠❛ ❞✐✈✐sã♦ ♣♦r ③❡r♦✳

x2 − 2x = x(x − 2) = 0✳

▼❛s q✉❛♥❞♦

q✉❡ ❞♦♠✐♥❛♠✱ ❡ ❛ ♠❡s♠❛ ❝♦❧♦❝❛çã♦ ❡♠ ❡✈✐❞ê♥❝✐❛ ♣♦❞❡ s❡r ❢❡✐t❛✱ ❡ ❝♦♥❝❧✉✐✲s❡ q✉❡

y = 1 é ❛ssí♥t♦t❛ ❤♦r✐③♦♥t❛❧ ✭❛ ❡sq✉❡r❞❛✮✳

❆ss✐♠✱ ♦s ❝❛♥❞✐❞❛t♦s sã♦

x=0

❡