Question

Evaluate using L'Hopital's rule: %29%5E%7Blnx%7D" id="TexFormula1" title="\lim_{x \to 1^{+}} (x - 1)^{lnx}" alt="\lim_{x \to 1^{+}} (x - 1)^{lnx}" align="absmiddle" class="latex-formula">

Answer 1

Answer:

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

Step-by-step 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:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{\ln(x-1)}{\ln(x)^{-1}}$

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

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}$

Simplify:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}\Big(\frac{-x\ln(x)^2}{-x\ln(x)^2}\Big)$

Simplify:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-\frac{x\ln(x)^2}{x-1}$

Now, by using direct substitution, we will acquire:

$\displaystyle \Rightarrow -\frac{1\ln(1)^2}{1-1}=\frac{0}{0}$

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:

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