What is the limit as x approaches 0 of 1/(sin^2(2x)) ?

The limit as x approaches 0 of 1/(sin^2(2x)) is infinity

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limit as x approaches 0 of 1/(sin^2(2x))

x \to 0 lim (\frac{1}{sin ^{2} ( 2 x )})

\infty

Solution steps

x \to 0 lim (\frac{1}{sin ^{2} ( 2 x )})

x \to a - lim f (x) = x \to a + lim f (x) = L

then

x \to a lim f (x) = L

x \to 0 - lim (\frac{1}{sin ^{2} ( 2 x )}) = \infty

x \to 0 + lim (\frac{1}{sin ^{2} ( 2 x )}) = \infty

= \infty

y^{^{'}} = t y + t

x y^{^{'}} + y = - 5 x y^{2}

\frac{d}{d x} ((2 x - 4) (- 3 x^{2} + 1))

x \to - 5 lim (\frac{5 - ∣ x ∣}{5 + x})

\int tan^{15} (x) sec^{4} (x) d x

What is the limit as x approaches 0 of 1/(sin^2(2x)) ?
The limit as x approaches 0 of 1/(sin^2(2x)) is infinity