Why does limx / 1-cosx = infinity instead of equal to 1?

Using equivalent infinitesimal to replace 1-cosx to (1 / 2) ^ 2

limx→0 (cosx+xsinx)
limx→0 (cosx+xsinx)^(1/x^2)

The original formula = limx → 0 (1 + xtanx) ^ (1 / x ^ 2) (cosx) ^ (1 / x ^ 2) = limx → 0 (1 + x ^ 2) ^ (1 / x ^ 2) (1 + cosx-1) ^ {[1 / (cosx-1)] [(cosx-1) / x ^ 2]} = elimx → 0e ^ [(cosx-1) / x ^ 2] limx → 0e ^ [(cosx-1) / x ^ 2] = limx → 0 (- SiNx) / 2x = - 1 / 2, so, the original formula = exe ^ (- 1 / 2

What is the limit of limx → π / 2 (2 + SiNx) / cosx

Because limx → π / 2, cosx / (2 + SiNx) = 0 / (2 + 1) = 0
therefore
The original formula = ∞

The value of the power of SiNx of limx approaching zero x

lim(x→0)x^sinx
=lim(x→0)e^(sinxlnx)
=lim(x→0)e^(xlnx)
=lim(x→0)e^(lnx/x^-1)
=lim(x→0)e^(-1/x/x^(-2))
=lim(x→0)e^(-x)
=1

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Why does limx / 1-cosx = infinity instead of equal to 1?