What is the limit as x approaches 0 of ((e^{2x}-e^{-2x}-4x))/((x-sin(x))) ?

The limit as x approaches 0 of ((e^{2x}-e^{-2x}-4x))/((x-sin(x))) is 16

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limit as x approaches 0 of ((e^{2x}-e^{-2x}-4x))/((x-sin(x)))

x \to 0 lim (\frac{( e ^{2 x} - e ^{- 2 x} - 4 x )}{( x - sin ( x ) )})

16

Solution steps

x \to 0 lim (\frac{e ^{2 x} - e ^{- 2 x} - 4 x}{x - sin ( x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2 e ^{2 x} + 2 e ^{- 2 x} - 4}{1 - cos ( x )})

= x \to 0 lim (\frac{2 e ^{2 x} + 2 e ^{- 2 x} - 4}{1 - cos ( x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 4 e ^{- 2 x} + 4 e ^{2 x}}{sin ( x )})

= x \to 0 lim (\frac{- 4 e ^{- 2 x} + 4 e ^{2 x}}{sin ( x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{8 e ^{- 2 x} + 8 e ^{2 x}}{cos ( x )})

= x \to 0 lim (\frac{8 e ^{- 2 x} + 8 e ^{2 x}}{cos ( x )})

Plug in the value

x = 0

= \frac{8 e ^{- 2 \cdot 0} + 8 e ^{2 \cdot 0}}{cos ( 0 )}

Simplify

\frac{8 e ^{- 2 \cdot 0} + 8 e ^{2 \cdot 0}}{cos ( 0 )} : 16

= 16

x \to 3 lim (x^{2} - 2 x + 7)

x \to 0 + lim (ln (x) cot (x))

x \to 1 lim (\frac{x + 2}{2})

n \to \infty lim (n^{2} - 1)

x \to - 4 - lim (x^{2} - 16)

What is the limit as x approaches 0 of ((e^{2x}-e^{-2x}-4x))/((x-sin(x))) ?
The limit as x approaches 0 of ((e^{2x}-e^{-2x}-4x))/((x-sin(x))) is 16