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

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

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

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

diverges

Solution steps

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

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

then the limit does not exist

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

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

= diverges

x \to 3 π lim (x^{3} sin^{4} (x))

x \to 0 lim (\frac{3 x ^{2} - 6 x}{x})

\frac{\partial}{\partial x} (7 x^{2} y + 5 x y^{3})

n = 1 \sum \infty (\frac{n}{n + 1})^{2}

x e^{- 0.2 x}, y = 0, x = 1

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