What is the limit as x approaches 0 of [ 1/x-1/(ln(1+x))] ?

The limit as x approaches 0 of [ 1/x-1/(ln(1+x))] is -1/2

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limit as x approaches 0 of [ 1/x-1/(ln(1+x))]

x \to 0 lim ([\frac{1}{x} - \frac{1}{ln ( 1 + x )}])

- \frac{1}{2}

Solution steps

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

Simplify

\frac{1}{x} - \frac{1}{ln ( 1 + x )} : \frac{ln ( x + 1 ) - x}{x ln ( x + 1 )}

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{- \frac{1}{( 0 + 1 ) ^{2}}}{\frac{0 + 2}{( 0 + 1 ) ^{2}}}

Simplify

\frac{- \frac{1}{( 0 + 1 ) ^{2}}}{\frac{0 + 2}{( 0 + 1 ) ^{2}}} : - \frac{1}{2}

= - \frac{1}{2}

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

x \to 0 lim (\frac{sin ( x ) + x}{x})

x \to 0 + lim (2 x csc (2 π - x))

x \to 2 lim (4 - 2 x)

x \to 0 lim (- \frac{1}{∣ x ∣})

What is the limit as x approaches 0 of [ 1/x-1/(ln(1+x))] ?
The limit as x approaches 0 of [ 1/x-1/(ln(1+x))] is -1/2