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

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

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

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

0

Solution steps

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

Express with sin, cos

= 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 )}

= 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 )})

= 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 )})

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

Plug in the value

x = 0

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

Simplify

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

= 0

\int_{0}^{π} 9 x^{2} sin (\frac{1}{2} x) d x

\frac{\partial}{\partial x} (- \frac{1}{x ^{2}} + e^{- y})

\frac{d y}{d x} = y (1 + x)^{- 1}

\int \frac{1}{( x + 3 ) ( x + 5 )} d x

\frac{\partial}{\partial x} ((5 x - x^{2}) cos (t))

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