Question

Evaluate using L'Hopital's rule:


`\lim_(x \to 1^(+)) (x - 1)^(lnx)`

Answer 1

`\displaystyle \lim_(x\to 1^+)(x-1)^\ln(x)}=1`

Explanation:

We are given:

`\displaystyle \lim_(x\to 1^+)(x-1)^\ln(x)`

And we want to evaluate it using L'Hopital's Rule.

First, using direct substitution, we will acquire:

`=(1-1)^\ln(1)}=0^0`

Which is indeterminate.

In order to apply L'Hopital's Rule, we first need to manipulate the expression. We will let:

`y=(x-1)^\ln(x)`

By taking the natural log of both sides:

`\ln(y)=\ln(x)\ln(x-1)`

And by taking the limit as x approaches 1 from the right of both functions:

`\displaystyle \lim_(x\to 1^+)\ln(y)=\lim_(x\to 1^+)\ln(x)\ln(x-1)`

Rewrite:

`\displaystyle \lim_(x\to 1^+)\ln(y)= \lim_(x\to 1^+)(\ln(x-1))/(\ln(x)^(-1))`

Using direct substitution on the right will result in 0/0. Hence, we can now apply L'Hopital's Rule:

`\displaystyle \lim_(x\to 1^+)\ln(y)= \lim_(x\to 1^+)(1/(x-1))/(-\ln(x)^(-2)(1/x))`

Simplify:

`\displaystyle \lim_(x\to 1^+)\ln(y)= \lim_(x\to 1^+)(1/(x-1))/(-\ln(x)^(-2)(1/x))\Big((-x\ln(x)^2)/(-x\ln(x)^2)\Big)`

Simplify:

`\displaystyle \lim_(x\to 1^+)\ln(y)= \lim_(x\to 1^+)-(x\ln(x)^2)/(x-1)`

Now, by using direct substitution, we will acquire:

`\displaystyle \Rightarrow -(1\ln(1)^2)/(1-1)=(0)/(0)`

Hence, we will apply L'Hopital's Rule once more. Utilize the product rule:

`\displaystyle \lim_(x\to 1^+)\ln(y)= \lim_(x\to 1^+)-(\ln(x)^2+2x\ln(x))`

Finally, direct substitution yields:

`\Rightarrow -(\ln(1)^2+2(1)\ln(1))=-(0+0)=0`

Thus:

`\displaystyle \lim_(x\to 1^+)\ln(y)=0`

By the Composite Function Property for limits:

`\displaystyle \lim_(x\to 1^+)\ln(y)=\ln( \lim_(x\to 1^+)y)=0`

Raising both sides to e produces:

`\displaystyle e^{\ln \lim_(x\to 1^+)y}=e^0`

Therefore:

`\displaystyle \lim_(x\to 1^+)y=1`

Substitution:

`\displaystyle \lim_(x\to 1^+)(x-1)^\ln(x)}=1`