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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

\int 12 x sec^{2} (x) d x

\frac{\partial}{\partial y} ((x + 2 y^{3})^{\frac{1}{3}})

(x^{4} + 3 x)^{- 1}

\int_{0}^{\frac{π}{2}} 1 d θ

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