Chapter 9.2, Problem 23E

To determine

To Find: The value of a limit using L'Hôpital's Rule, after identifying the indeterminate form.

Answer to Problem 23E

The value of the limit is $1$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 1}{\lim }{\left(x^2-2x+1\right)}^{x-1}$ .

Method: $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(x)\text{ and }g(x)\text{ approach infinity as }x\to c,\text{ then }\underset{x\to c}{\lim }\frac{f(x)}{g(x)}=\underset{x\to c}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: As $a\to 1$ ,

  $a^2-2a+1\to 0$ and $a-1\to 0$

Therefore, the limit is of the indeterminate form $0^0$ .

So, assume the limit is $L$ and take the natural logarithm of the equation $L=\underset{x\to 1}{\lim }{\left(x^2-2x+1\right)}^{x-1}$ so that

  $\begin{array}{l}L=\underset{x\to 1}{\lim }{\left(x^2-2x+1\right)}^{x-1}\\ \Rightarrow \ln L=\ln \left[\underset{x\to 1}{\lim }{\left(x^2-2x+1\right)}^{x-1}\right]\\ \Rightarrow \ln L=\underset{x\to 1}{\lim }\left[\ln {\left(x^2-2x+1\right)}^{x-1}\right]\\ \Rightarrow \ln L=\underset{x\to 1}{\lim }\left[\left(x-1\right)\ln \left(x^2-2x+1\right)\right]\\ \Rightarrow \ln L=\underset{x\to 1}{\lim }\left[\frac{\ln \left(x^2-2x+1\right)}{\frac{1}{x-1}}\right]\end{array}$

Now, the righthand side of the above equation is in $\frac{\infty }{\infty }$ form.

So, apply L'Hôpital's Rule on the righthand side of the equation to write

  $\begin{array}{l}\ln L=\underset{x\to 1}{\lim }\frac{\left[\raisebox{1ex}{$2x-2+0$}\!\left/ \!\raisebox{-1ex}{$x^2-2x+1$}\right.\right]}{\left[\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{${(x-1)}^2$}\right.\right]}\\ =\underset{x\to 1}{\lim }\frac{\left[\raisebox{1ex}{$2(x-1)$}\!\left/ \!\raisebox{-1ex}{${(x-1)}^2$}\right.\right]}{\left[\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{${(x-1)}^2$}\right.\right]}\\ =\underset{x\to 1}{\lim }\left[-2(x-1)\right]\\ =-2(1-1)\\ =0\\ \Rightarrow L=e^0=1\end{array}$

Therefore, the value of the limit is $1$ .