Chapter 9.2, Problem 40E

To determine

To Find: The value of a limit using L'Hôpital's Rule.

Answer to Problem 40E

The value of the limit is $1$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0}{\lim }{\left(\frac{1}{x^2}\right)}^x$ .

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

Calculation: As $x\to 0$ ,

  $\frac{1}{x^2}\to \infty$ and $x\to 0$

Therefore, the limit is of the indeterminate form ${\infty }^0$ . So, assume the value of the limit is $L$ and then take the natural logarithm of the limit to convert it into $\frac{\infty }{\infty }$ form as below.

  $L=\underset{x\to 0}{\lim }{\left(\frac{1}{x^2}\right)}^x$

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

Then, replace $\frac{1}{x}$ with $h$ to write

  $\Rightarrow \ln L=\underset{x\to 0}{\lim }\frac{\ln \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.\right)}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}=\underset{h\to \infty }{\lim }\frac{\ln h^2}{h}=\underset{h\to \infty }{\lim }\frac{2\ln h}{h}$

Now, apply L'Hôpital's Rule to write

  $\underset{h\to \infty }{\lim }\frac{2\ln h}{h}=\underset{h\to \infty }{\lim }\frac{2\cdot \frac{1}{h}}{1}=2\cdot 0=0$

  $\Rightarrow L=e^0=1$

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