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

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

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

x \to 1 + lim ((x - 1) ln (x - 1))

0

Solution steps

x \to 1 + lim ((x - 1) ln (x - 1))

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

= x \to 1 + lim (1 - x)

Plug in the value

x = 1

= 1 - 1

Simplify

= 0

x \to \infty lim (3 27 + \frac{6}{x} - 3)

x \to \infty lim (\frac{cosh ( 2 x )}{e ^{2 x}})

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

x \to 0 lim (arctan (\frac{3}{x}))

x \to 0 - lim (x^{3} sin (\frac{π}{x}))

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