limx->∞ x²(1-cos(2π/x)) How to calculate this? ∞-∞

Your idea is wrong. You should regard it as ∞ * 0. Then you can change it to 0 / 0 and use the lobita rule

The limit of x-in (x + 1) / X & # can be found by the law of lobita

X tends to 0. Obviously, ln (x + 1) and X & # 178; tend to 0 when X - > 0, which satisfies the condition of lobita's rule. Therefore, the derivation of molecular denominator is carried out at the same time. Obviously [x-in (x + 1)] '= 1 - 1 / (x + 1) = x / (x + 1), and (X & # 178;)' = 2x, so LIM (x - > 0) [x-in (x + 1)] / X & # 178; =

The limit of (1-x) Tan (π X / 2) x → 1
Find the limit of (1-x) Tan (π X / 2) and X → 1

Tan (π X / 2) = 1 / Tan (1-x) π / 2 ~ 1 / (1-x) π / 2
So (1-x) Tan (π X / 2) = (1-x) / (1-x) π / 2 = 2 / π
The limit sign is omitted here for convenience

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limx->∞ x²(1-cos(2π/x)) How to calculate this? ∞-∞