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

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

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

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

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 1

= \frac{- \frac{1}{1 ^{2}}}{\frac{1}{1 ^{2}} + \frac{1}{1}}

Simplify

\frac{- \frac{1}{1 ^{2}}}{\frac{1}{1 ^{2}} + \frac{1}{1}} : - \frac{1}{2}

= - \frac{1}{2}

\int \frac{( e ^{6 x} )}{e ^{6 x} + 1} d x

y = x, y = 2 x 64 - x^{2}

\frac{d y}{d x} = y^{2} e^{3 x}

\frac{\partial ^{2}}{\partial r \partial θ} (r (θ) \cdot cos (θ))

x = 6 y^{2}, x + y = 5

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