What is the limit as x approaches 0-of 4(1/x-csc(x)) ?

The limit as x approaches 0-of 4(1/x-csc(x)) is 0

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limit as x approaches 0-of 4(1/x-csc(x))

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

0

Solution steps

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

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

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

Express with sin, cos

= 4 \cdot x \to 0 - lim (\frac{1}{x} - \frac{1}{sin ( x )})

Simplify

\frac{1}{x} - \frac{1}{sin ( x )} : \frac{sin ( x ) - x}{x sin ( x )}

= 4 \cdot x \to 0 - lim (\frac{sin ( x ) - x}{x sin ( x )})

Apply L'Hopital's Rule

: x \to 0 - lim (\frac{cos ( x ) - 1}{sin ( x ) + x cos ( x )})

= 4 \cdot x \to 0 - lim (\frac{cos ( x ) - 1}{sin ( x ) + x cos ( x )})

Apply L'Hopital's Rule

: x \to 0 - lim (\frac{- sin ( x )}{2 cos ( x ) - x sin ( x )})

= 4 \cdot x \to 0 - lim (\frac{- sin ( x )}{2 cos ( x ) - x sin ( x )})

Plug in the value

x = 0

= 4 \cdot \frac{- sin ( 0 )}{2 cos ( 0 ) - 0 \cdot sin ( 0 )}

Simplify

4 \cdot \frac{- sin ( 0 )}{2 cos ( 0 ) - 0 \cdot sin ( 0 )} : 0

= 0

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

\frac{\partial}{\partial x} (2 y - x^{2} e^{- y})

t^{3} e^{b \cdot t}

\frac{\partial}{\partial x} (ln (x + y^{4}))

x \to \infty lim (- \frac{x}{2})

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