How to find LIM (x → 1) (3x + 5) / (x ^ 3-1)?

How to find LIM (x → 1) (3x + 5) / (x ^ 3-1)?

(3x+5)/(x^3-1)=(3X+5)/(X-1)(X^2+X+1)
=3(X-1)/(X-1)(X^2+X+1)+8/(X^3-1)
=3/[(X^2+X+1/4)+2/4]+8/(X^3-1)
=3/[(X+1/2)^2+3/4] + 8/(X^3-1)
Let x = 1, then the original formula = 1 + ∞ = ∞

The limit of LIM (x3-3x + 2) / (x4-4x + 3) when x tends to 1

LIM (x3-3x + 2) / (x4-4x + 3) = LIM (x ^ 3-x ^ 2 + x ^ 2-3x + 2) / (x ^ 4-x ^ 2 + x ^ 2-4x + 3) (about x-1)
=LIM (x ^ 2 + X-2) / (x ^ 3 + x ^ 2 + x-3) (about x-1)
=lim（x＋2）/（x^2+2x＋3）=3/6=1/2,

LIM (x → π / 2) ln (1 + cosx) / (π / 2-x) the answer is 1
LIM (x → π / 2) (LN (1 + cosx)) / (π / 2-x) the answer is 1

Infinity
=1+0/0=1/0
That's wrong, X → π / 2, cosx -- 0, (π / 2-x -- 0)
1 / 0, unless you have the wrong title