What is the limit as x approaches 0 of 1/(2xcsc(4x)) ?

The limit as x approaches 0 of 1/(2xcsc(4x)) is 2

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limit as x approaches 0 of 1/(2xcsc(4x))

x \to 0 lim (\frac{1}{2 x csc ( 4 x )})

2

Solution steps

x \to 0 lim (\frac{1}{2 x csc ( 4 x )})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{2} \cdot x \to 0 lim (\frac{1}{x csc ( 4 x )})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{1}{2} \cdot \frac{lim _{x \to 0} ( 1 )}{lim _{x \to 0} ( x csc ( 4 x ) )}

x \to 0 lim (1) = 1

x \to 0 lim (x csc (4 x)) = \frac{1}{4}

= \frac{1}{2} \cdot \frac{1}{\frac{1}{4}}

Simplify

\frac{1}{2} \cdot \frac{1}{\frac{1}{4}} : 2

= 2

x \to 0 + lim ((\frac{x}{34})^{x})

\int_{0}^{16} \frac{1}{64 - x ^{2}} d x

y^{3} - (14 x + 4) + 3 x y^{2} y^{^{'}} = 0

y + y^{4} = (y^{6} + 3 x) \frac{d y}{d x}

y = (\frac{( 3 x ^{2} )}{( 2 x + 4 )})^{3}

What is the limit as x approaches 0 of 1/(2xcsc(4x)) ?
The limit as x approaches 0 of 1/(2xcsc(4x)) is 2