Mathematical limit LIM (x → 0) ((x ^ 3) / (3 * x ^ 2-1)) LIM (x → + infinity) (√ x (x + 2) - √ (x ^ 2-x + 1)) LIM (x → 1) (Tan (x-1)) / (x ^ 2-1)

Mathematical limit LIM (x → 0) ((x ^ 3) / (3 * x ^ 2-1)) LIM (x → + infinity) (√ x (x + 2) - √ (x ^ 2-x + 1)) LIM (x → 1) (Tan (x-1)) / (x ^ 2-1)

lim(x→0)((x^3)/(3*x^2-1))
=lim(x→0)3x^2/6x=0
lim(x→+∞)(√x(x+2)-√(x^2-x+1)) =lim(x→+∞)(√(x+1)^2-1-√x-1/2)^2+3/4)
= lim(x→+∞)(x+1-(x-1/2))
=3/2
lim(x→1)(tan(x-1))/(x^2-1)
=lim(x→1)(sin(x-1)/cos(x-1)(x^2-1)
=LIM (x → 1) (sin (x-1) / cos (x-1) (x-1) (x + 1) when X-1 → 0, limsinx = limx
So LIM (x → 1) (sin (x-1) / cos (x-1) (x-1) (x + 1)
=lim(x→1)1/cos(x-1)(x+1)
=1/2

Find LIM (x - > + infinity) (2 / arctan x) ^ x
The PI school hopes to understand it. Please give me a specific solution,

The original formula = Lim [1 + (2arctanx - π) / π] ^ [π / (2arctanx - π) * (2arctanx - π) / π * x] = e ^ LIM (2arctanx - π) / (π / x) = e ^ Lim [2 / (1 + x ^ 2)] / (- 1 / x ^ 2) = e ^ (- 2) limx ^ 2 / (1 + x ^ 2) = e ^ (- 2) Lim [1-1 / (1 + x ^ 2)] = e ^ (- 2)

Solving LIM (x → + ∞) (π / 2 - arctan * (2x ^ 2)) x ^ 2

When x tends to + ∞, arctan (2x ^ 2) tends to π / 2,
So π / 2-arctan (2x ^ 2) tends to zero,
And x ^ 2 tends to infinity,
therefore
Original limit = LIM (x tends to + ∞) [π / 2-arctan (2x ^ 2)] / (1 / x ^ 2)
At this time, the numerator denominator tends to zero, which satisfies the application conditions of lobita's law, and the numerator denominator is derived at the same time,
that
Original limit
=LIM (x tends to + ∞) [- 4x / (1 + 4x ^ 4)] / (- 2 / x ^ 3)
=LIM (x tends to + ∞) 2x ^ 4 / (1 + 4x ^ 4)
=LIM (x tends to + ∞) 2 / (1 / x ^ 4 + 4)
Obviously, when x tends to + ∞, 1 / x ^ 4 tends to 0,
so
Original limit = 2 / 4 = 1 / 2

Lim(x→a)arctan(x-a)/(x-a)

Let t = x-a
Original formula = LIM (T - > 0) arctant / T
From the law of lobida
=lim 1/1+t^2
=1