Are you allowed to use L'Hopital's rule?

yes!

1a) is quite easy.

Iim (x² - 9) / (x² - 7x + 12)
x→3

= Iim (x + 3)(x - 3) / [(x - 3)(x - 4)]
x→3

= Iim (x + 3) / (x - 4)
x→3

= (3 + 3) / (3 - 4)

= 6 / -1

= -6

1b)

lim (2x³ - 1458) / (√x - 3)
x→9

= lim 2(x³ - 729) / (√x - 3)
x→9

= 2 lim (x³ - 9³) / (√x - 3)
x→9

= 2 lim (x - 9)(x² + x(9) + 9²) / (√x - 3)
x→9

[Recall the property a³ - b³ = (a - b)(a² + ab + b²) ]

= 2 lim (x - 9)(x² + 9x + 81) / (√x - 3)
x→9

= 2 lim ((√x)² - 3²)(x² + 9x + 81) / (√x - 3)
x→9

= 2 lim (√x - 3)(√x + 3)(x² + 9x + 81) / (√x - 3)
x→9

[Recall the property a² - b² = (a + b)(a - b) ]

= 2 lim (√x + 3)(x² + 9x + 81)
x→9

= 2 (√9 + 3)(9² + 9(9) + 81)

= 2(6)(243)

= 2916

(Edited)

1c)

lim x⁴ sin π/x cos π/x²
x→0

= lim (x⁴) · lim (sin π/x cos π/x²)
x→0□□□x→0

0⁴ · lim (sin π/x cos π/x²))
□□x→0

= 0 · lim sin π/x cos π/x²
□□x→0

= 0

(0 multiplied by anything is still 0)


(Note that if we were to find lim sin π/x cos π/x², x→0 alone, the limit does not exist )

2a) (using L'Hopital's rule)

lim ((2x + 4)³/² + e²ˣ - 9) / 2x
x→0

= lim d/dx((2x + 4)³/² + e²ˣ - 9) / d/dx(2x)
x→0

= lim (3/2 ·(2x + 4)¹/² · 2 + 2e²ˣ) / 2
x→0

= lim (3/2 · (2x + 4)¹/² + e²ˣ)
x→0

= 3/2 · (2(0) + 4)¹/² + e²⁽⁰⁾

= 3/2 · 4¹/² + 1

= 3/2 · 2 + 1

= 3 + 1

= 4

2a) (using Taylor/Maclaurin Series)

Using the series, you will be able to derive the following :

(2x + 4)³/² = 8 + 6x + ¾ x² - 1/16 x³ + 3/256 x⁴ + ...

e²ˣ = 1 + 2x + 2x² + 4/3 x³ + ⅔ x⁴ + ...


So,

((2x + 4)³/² + e²ˣ + 9) / 2x

= ( (8 + 6x + ¾ x² - 1/16 x³ + 3/256 x⁴ + ...) + (1 + 2x + 2x² + 4/3 x³ + ⅔ x⁴ + ...) - 9) / 2x

= ((6x + ¾ x² - 1/16 x³ + 3/256 x⁴ + ...) + (2x + 2x² + 4/3 x³ + ⅔ x⁴ + ...) + 8 + 1 - 9) / 2x

= ((6x + ¾ x² - 1/16 x³ + 3/256 x⁴ + ...) + (2x + 2x² + 4/3 x³ + ⅔ x⁴ + ...)) / 2x

= (3 + ⅜ x - 1/32 x² + 3/512 x³ + ...) + (1 + x + ⅔ x² + ⅓ x³ + ...)

= 3 + 1 + (⅜ x - 1/32 x² + 3/512 x⁴ + ...) + (x + ⅔ x² + ⅓ x³ + ...)

= 4 + (⅜ x - 1/32 x² + 3/512 x⁴ + ...) + (x + ⅔ x² + ⅓ x³ + ...)


Then,

lim ((2x + 4)³/² + e²ˣ + 9) / 2x
x → 0

= lim (4 + (⅜ x - 1/32 x² + 3/512 x⁴ + ...) + (x + ⅔ x² + ⅓ x³ + ...) )
x → 0

= 4 + (⅜ (0) - 1/32 (0²) + 3/512 (0⁴) + ...) + (0 + ⅔ (0²) + ⅓ (0³) + ...)

= 4