What is the limit as x approaches 0 of (ln(\frac{1+x)/(1-x))-2x}{x-sin(x)} ?

The limit as x approaches 0 of (ln(\frac{1+x)/(1-x))-2x}{x-sin(x)} is 4

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limit as x approaches 0 of (ln(\frac{1+x)/(1-x))-2x}{x-sin(x)}

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

4

Solution steps

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

Apply L'Hopital's Rule

: x \to 0 lim (\frac{4}{cos ( x ) ( x + 1 ) ^{2} ( - x + 1 ) ^{2} + ( 4 x ^{3} - 4 x ) sin ( x )})

= x \to 0 lim (\frac{4}{cos ( x ) ( x + 1 ) ^{2} ( - x + 1 ) ^{2} + ( 4 x ^{3} - 4 x ) sin ( x )})

Plug in the value

x = 0

= \frac{4}{cos ( 0 ) ( 0 + 1 ) ^{2} ( - 0 + 1 ) ^{2} + ( 4 \cdot 0 ^{3} - 4 \cdot 0 ) sin ( 0 )}

Simplify

\frac{4}{cos ( 0 ) ( 0 + 1 ) ^{2} ( - 0 + 1 ) ^{2} + ( 4 \cdot 0 ^{3} - 4 \cdot 0 ) sin ( 0 )} : 4

= 4

\int x x^{4} + 1 d x

f (x) = 6 x - x^{2}, at x = 0

f (t) = 80 t + t^{2} - \frac{1}{48} t^{3}

- 4 x + x^{2} + 10, 12 - 5 x, 4 x - 6

\int_{1}^{16} \frac{8}{x} d x

What is the limit as x approaches 0 of (ln(\frac{1+x)/(1-x))-2x}{x-sin(x)} ?
The limit as x approaches 0 of (ln(\frac{1+x)/(1-x))-2x}{x-sin(x)} is 4