Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 58

Determine the $\displaystyle \lim_{x \to \infty} x \frac{\ln 2}{1 + \ln x}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

If we let $\displaystyle y = x \frac{\ln 2}{1 + \ln x}$, then
$\displaystyle \ln y = \left( \frac{\ln 2}{1 + \ln x} \right) \ln x$

So,
$\displaystyle \lim_{x \to \infty} \ln y = \lim_{x \to \infty} \left( \frac{\ln 2 (\ln x)}{1 + \ln x} \right)$

By applying L'Hospital's Rule...

$
\begin{equation}
\begin{aligned}
\lim_{x \to \infty} \left( \frac{\ln 2 (\ln x)}{1 + \ln x} \right) &= \lim_{x \to \infty} \frac{\ln 2 \left( \frac{1}{x} \right)}{0 + \left( \frac{1}{x} \right)}\\
\\
&= \lim_{x \to \infty} \ln 2\\
\\
&= \ln 2
\end{aligned}
\end{equation}
$


Thus,

$\displaystyle \lim_{x \to \infty} \ln y = \lim_{x \to \infty} \left( \frac{\ln 2(\ln x)}{1 + \ln x} \right) = \ln 2$
Therefore, we have
$\displaystyle \lim_{x \to \infty} x \frac{\ln 2}{1 + \ln x} = \lim_{x \to \infty} e^{\ln y} = e^{\ln 2} = 2$