What is the limit as x approaches 0+of 6(cot(x)-1/x) ?

The limit as x approaches 0+of 6(cot(x)-1/x) is 0

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limit as x approaches 0+of 6(cot(x)-1/x)

x \to 0 + lim (6 (cot (x) - \frac{1}{x}))

0

Solution steps

x \to 0 + lim (6 (cot (x) - \frac{1}{x}))

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

= 6 \cdot x \to 0 + lim (cot (x) - \frac{1}{x})

Express with sin, cos

= 6 \cdot 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 )}

= 6 \cdot 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 )})

= 6 \cdot 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 )})

= 6 \cdot x \to 0 + lim (\frac{- sin ( x ) - x cos ( x )}{2 cos ( x ) - x sin ( x )})

Plug in the value

x = 0

= 6 \cdot \frac{- sin ( 0 ) - 0 \cdot cos ( 0 )}{2 cos ( 0 ) - 0 \cdot sin ( 0 )}

Simplify

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

= 0

\frac{d}{d x} (- e^{- 2 x})

f (x) = 5 x + 4, at x = 9

y^{^{''}} + 10 y^{^{'}} + 25 y = 2 r^{- 5 x}

\int ln (x^{\frac{1}{3}}) d x

\int (3 - cos^{3} (x)) sin (x) d x

What is the limit as x approaches 0+of 6(cot(x)-1/x) ?
The limit as x approaches 0+of 6(cot(x)-1/x) is 0