Answer:
[tex]\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1[/tex]
Step-by-step explanation:
We are given:
[tex]\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)[/tex]
And we want to evaluate it using L'Hopital's Rule.
First, using direct substitution, we will acquire:
[tex]=(1-1)^\ln(1)}=0^0[/tex]
Which is indeterminate.
In order to apply L'Hopital's Rule, we first need to manipulate the expression. We will let:
[tex]y=(x-1)^\ln(x)[/tex]
By taking the natural log of both sides:
[tex]\ln(y)=\ln(x)\ln(x-1)[/tex]
And by taking the limit as x approaches 1 from the right of both functions:
[tex]\displaystyle \lim_{x\to 1^+}\ln(y)=\lim_{x\to 1^+}\ln(x)\ln(x-1)[/tex]
Rewrite:
[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{\ln(x-1)}{\ln(x)^{-1}}[/tex]
Using direct substitution on the right will result in 0/0. Hence, we can now apply L'Hopital's Rule:
[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}[/tex]
Simplify:
[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}\Big(\frac{-x\ln(x)^2}{-x\ln(x)^2}\Big)[/tex]
Simplify:
[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-\frac{x\ln(x)^2}{x-1}[/tex]
Now, by using direct substitution, we will acquire:
[tex]\displaystyle \Rightarrow -\frac{1\ln(1)^2}{1-1}=\frac{0}{0}[/tex]
Hence, we will apply L'Hopital's Rule once more. Utilize the product rule:
[tex]\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-(\ln(x)^2+2x\ln(x))[/tex]
Finally, direct substitution yields:
[tex]\Rightarrow -(\ln(1)^2+2(1)\ln(1))=-(0+0)=0[/tex]
Thus:
[tex]\displaystyle \lim_{x\to 1^+}\ln(y)=0[/tex]
By the Composite Function Property for limits:
[tex]\displaystyle \lim_{x\to 1^+}\ln(y)=\ln( \lim_{x\to 1^+}y)=0[/tex]
Raising both sides to e produces:
[tex]\displaystyle e^{\ln \lim_{x\to 1^+}y}=e^0[/tex]
Therefore:
[tex]\displaystyle \lim_{x\to 1^+}y=1[/tex]
Substitution:
[tex]\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1[/tex]
