limits

[halcyon lodge]

need to calculate it using vanishing functions ( idk if i translated it right)

[bold mica]

$\frac{cos^3(x)-1}{sin^2(2x)}\$

$\= \frac{cos^3(x)-1^3}{(2sin(x)cos(x))^2} \$

$\=\frac{(cos(x)-1) \cdot (cos^2(x) +cos(x)+1)}{4sin^2(x)cos^2(x)}\$

$\= \frac{-(1-cos(x)) \cdot (cos^2(x) +cos(x)+1)}{4(1^2-cos^2(x))cos^2(x)}\$

$\= \frac{-(1-cos(x)) \cdot (cos^2(x) +cos(x)+1)}{4(1-cos(x))(1+cos(x))cos^2(x)}\$

$\= -\frac{cos^2(x) +cos(x)+1}{4cos^2(x)+4cos^3(x)}$

[maiden basin]

adonhs

[bold mica]

$\-\frac{cos^2(0) +cos(0)+1}{4cos^2(0)+4cos^3(0)} = -\frac{3}{8}$

[maiden basin]

adonhs

[bold mica]

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