11
Calcula:
[
a) lím
x→2
(√
c) lím
x→0
a) lím
x→2
x2
[
x+9–3 x2
x2
= lím
x→2
3 4 – – 5x + 6 x – 2
)
]
b) lím
x→2
(
1 – √3 – x x–2
)
lím (2x + 1 – √4x 2 + 1 )
d)
x → +∞
]
[
]
3 4 3 4 – = lím – = (x – 2)(x – 3) (x – 2) – 5x + 6 x–2 x→2
3 – 4(x – 3) 3 – 4x + 12 –4x + 15 7 = lím = lím = (x – 2)(x – 3) (x – 2)(x – 3) (x – 2)(x – 3) (0) x→2 x→2
Hallamos los límites laterales: –4x + 15 = +∞; (x – 2)(x – 3)
lím
x→
2–
b) lím
x→2
lím
x→
2+
–4x + 15 = –∞ (x – 2)(x – 3)
— — (1 – √3 – x )(1 + √3 – x ) = 1 – √3 – x = lím x–2 x→2 (x – 2)(1 + √ 3 – x )
= lím
1 – (3 – x) 1–3+x = lím = x → 2 (x – 2)(1 + √ 3 – x ) (x – 2)(1 + √ 3 – x )
= lím
x–2 1 1 = lím = 2 x → 2 (x – 2)(1 + √ 3 – x ) 1 + √3 – x
x→2
x→2
√x + 9 – 3 = c) lím lím 2 x
x→0
= lím
x → 0 x2
—
—
(√x + 9 – 3)(√x + 9 + 3) x 2 (√ x + 9 + 3)
x→0
= lím
x→0
x+9–9 = x 2 (√ x + 9 + 3)
x 1 1 = lím = (0) (√ x + 9 + 3) x → 0 x (√ x + 9 + 3)
Hallamos los límites laterales: lím
x→
0–
1 = – ∞; x (√ x + 9 + 3)
1 = +∞ x (√ x + 9 + 3)
lím
x→
0+
—
d) lím
x → +∞
=
=
(2x + 1 –
√4x 2 + 1 ) =
lím
(2x + 1 + √ 4x 2 + 1)
x → +∞
lím
(2x + 1) 2 – (4x 2 + 1) = (2x + 1 + √ 4x 2 + 1)
lím
4x + 2 4 4 = = =1 2 2 + 2 4 (2x + 1 + √ 4x + 1)
x → +∞
x → +∞
Unidad 9. Límites de funciones. Continuidad
—
(2x + 1 – √4x 2 + 1)(2x + 1 + √4x 2 + 1)
lím
x → +∞
4x 2 + 4x + 1 – 4x 2 + 1 = (2x + 1 + √ 4x 2 + 1)
23
=