Chapter 9.2, Problem 43E

To determine

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

Answer to Problem 43E

The value of the limit is $\sqrt{e}$ .

Explanation of Solution

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

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 $x\to \infty$ ,

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

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.

  $L=\underset{x\to \infty }{\lim }{(1+2x)}^{\frac{1}{2\ln x}}$

  $\Rightarrow \ln L=\ln \left[\underset{x\to \infty }{\lim }{(1+2x)}^{\frac{1}{2\ln x}}\right]=\underset{x\to \infty }{\lim }\left[\ln {(1+2x)}^{\frac{1}{2\ln x}}\right]=\underset{x\to \infty }{\lim }\frac{\ln (1+2x)}{2\ln x}$

Now, apply L'Hôpital's Rule to write

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{\ln (1+2x)}{2\ln x}=\underset{x\to \infty }{\lim }\frac{\left[\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\left(1+2x\right)$}\right.\right]}{\left[2\cdot \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)\right]}=\frac{1}{2}\underset{x\to \infty }{\lim }\frac{2x}{1+2x}\\ =\frac{1}{2}\underset{x\to \infty }{\lim }\left(\frac{2\cdot 1}{0+2\cdot 1}\right)\left[\text{L'Hôpital's Rule applied again as limit is still in }\frac{\infty }{\infty }\text{ form}\right]\\ =\frac{1}{2}\end{array}$

  $\Rightarrow L=e^{\frac{1}{2}}=\sqrt{e}$

Therefore, the value of the limit is $\sqrt{e}$ .