Does anyone know how to solve this? It’s really starting to stress me out. PS: You just can solve the lim(x->2) instead of lim(u->0) if u want

π − 12

Step-by-step explanation:

lim(x→2) (sin(πx) + 8 − x³) / (x − 2)

If we substitute x = u + 2:

lim(u→0) (sin(π(u + 2)) + 8 − (u + 2)³) / ((u + 2) − 2)

lim(u→0) (sin(πu + 2π) + 8 − (u + 2)³) / u

Distribute the cube:

lim(u→0) (sin(πu + 2π) + 8 − (u³ + 6u² + 12u + 8)) / u

lim(u→0) (sin(πu + 2π) + 8 − u³ − 6u² − 12u − 8) / u

lim(u→0) (sin(πu + 2π) − u³ − 6u² − 12u) / u

Using angle sum formula:

lim(u→0) (sin(πu) cos(2π) + sin(2π) cos(πu) − u³ − 6u² − 12u) / u

lim(u→0) (sin(πu) − u³ − 6u² − 12u) / u

Divide:

lim(u→0) [ (sin(πu) / u) − u² − 6u − 12 ]

lim(u→0) (sin(πu) / u) + lim(u→0) (-u² − 6u − 12)

lim(u→0) (sin(πu) / u) − 12

Multiply and divide by π.

lim(u→0) (π sin(πu) / (πu)) − 12

π lim(u→0) (sin(πu) / (πu)) − 12

Use special identity, lim(x→0) ((sin x) / x ) = 1.

π (1) − 12

π − 12