What is the limit as x approaches 0 of (e^x-e^{-x}-2ln(1+x))/(xsin(x)) ?

The limit as x approaches 0 of (e^x-e^{-x}-2ln(1+x))/(xsin(x)) is 1

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limit as x approaches 0 of (e^x-e^{-x}-2ln(1+x))/(xsin(x))

x \to 0 lim (\frac{e ^{x} - e ^{- x} - 2 ln ( 1 + x )}{x sin ( x )})

1

Solution steps

x \to 0 lim (\frac{e ^{x} - e ^{- x} - 2 ln ( 1 + x )}{x sin ( x )})

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{e ^{0} - e ^{- 0} + \frac{2}{( 1 + 0 ) ^{2}}}{2 cos ( 0 ) - 0 \cdot sin ( 0 )}

Simplify

\frac{e ^{0} - e ^{- 0} + \frac{2}{( 1 + 0 ) ^{2}}}{2 cos ( 0 ) - 0 \cdot sin ( 0 )} : 1

= 1

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

x \to 10 lim ((x - 10)^{10})

x \to 2 lim (x^{2} - 5 x + 4)

x \to 0 lim (x^{n} e^{- x})

h \to 0 lim (\frac{1}{1 + h ^{2}})

What is the limit as x approaches 0 of (e^x-e^{-x}-2ln(1+x))/(xsin(x)) ?
The limit as x approaches 0 of (e^x-e^{-x}-2ln(1+x))/(xsin(x)) is 1