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

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

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limit as x approaches 0+of cot(x)-1/x

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

0

Solution steps

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

Express with sin, cos

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 0

\int 1 + 4 x d x

\frac{1}{s} \cdot \frac{1}{s + 1}

\int 8 sin^{2} (x + \frac{π}{12}) d x

n = 1 \sum \infty \infty

\int \frac{1}{x ^{\frac{3}{2}} + x ^{\frac{1}{2}}} d x

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