What is the limit as x approaches 0 of 7/x-7cot(x) ?

The limit as x approaches 0 of 7/x-7cot(x) is 0

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limit as x approaches 0 of 7/x-7cot(x)

x \to 0 lim (\frac{7}{x} - 7 cot (x))

0

Solution steps

x \to 0 lim (\frac{7}{x} - 7 cot (x))

Simplify

\frac{7}{x} - 7 cot (x) : \frac{7 - 7 x cot ( x )}{x}

= x \to 0 lim (\frac{7 - 7 x cot ( x )}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 7 ( - x csc ^{2} ( x ) + cot ( x ) )}{1})

= x \to 0 lim (\frac{- 7 ( - x csc ^{2} ( x ) + cot ( x ) )}{1})

Simplify

= x \to 0 lim (- 7 (- x csc^{2} (x) + cot (x)))

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

= - 7 \cdot x \to 0 lim (- x csc^{2} (x) + cot (x))

Express with sin, cos

= - 7 \cdot x \to 0 lim (- \frac{x}{sin ^{2} ( x )} + \frac{cos ( x )}{sin ( x )})

Simplify

- \frac{x}{sin ^{2} ( x )} + \frac{cos ( x )}{sin ( x )} : \frac{- x + cos ( x ) sin ( x )}{sin ^{2} ( x )}

= - 7 \cdot x \to 0 lim (\frac{- x + cos ( x ) sin ( x )}{sin ^{2} ( x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 1 + cos ( 2 x )}{sin ( 2 x )})

= - 7 \cdot x \to 0 lim (\frac{- 1 + cos ( 2 x )}{sin ( 2 x )})

Apply L'Hopital's Rule

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

= - 7 \cdot x \to 0 lim (\frac{- 2 sin ( 2 x )}{cos ( 2 x ) \cdot 2})

Plug in the value

x = 0

= - 7 \cdot \frac{- 2 sin ( 2 \cdot 0 )}{cos ( 2 \cdot 0 ) \cdot 2}

Simplify

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

= 0

f (x) = \frac{x}{7 + x}

f (x) = e^{x} - e^{- x} - 2 x

x \to - 1 lim (\frac{ln ( x )}{x})

5 H

\frac{d}{d x} (2 x^{\frac{5}{2}})

What is the limit as x approaches 0 of 7/x-7cot(x) ?
The limit as x approaches 0 of 7/x-7cot(x) is 0