Answer
$$
\lim _{x \rightarrow \infty}(x-\ln x)=\infty
$$
Work Step by Step
$$
\lim _{x \rightarrow \infty}(x-\ln x) =\infty
$$
First notice that
$$
\lim _{x \rightarrow \infty} x \,\,\, \rightarrow \infty
$$
and
$$
\lim _{x \rightarrow \infty }\ln x \,\,\, \rightarrow \infty ,
$$
so the limit is indeterminate and has the form $\infty -\infty $. we will change the form to a product by factoring out $x$:
$$
\begin{aligned}
\lim _{x \rightarrow \infty}(x-\ln x) & =\lim _{x \rightarrow \infty} x\left(1-\frac{\ln x}{x}\right) \\
& = \lim _{x \rightarrow \infty} x . \lim _{x \rightarrow \infty}\left(1-\frac{\ln x}{x}\right)\\
& =\lim _{x \rightarrow \infty} x . \left(1- \lim _{x \rightarrow \infty} \frac{\ln x}{x}\right)\\
& =\lim _{x \rightarrow \infty} x . \left(1- \lim _{x \rightarrow \infty} \frac{\ln x}{x}\right)\\
& \quad\quad [\text { since } \lim _{x \rightarrow \infty} \frac{\ln x}{x} \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow \infty} \frac{1 / x}{1}=0]\\
&=\lim _{x \rightarrow \infty} x (1-0) \\
&=\lim _{x \rightarrow \infty} x\\
&=\infty
\end{aligned}
$$
Note that the use of l’Hospital’s Rule is justified because:
$ x \rightarrow \infty \quad $ and $ \quad \ln x \rightarrow \infty $ as$ \quad x \rightarrow \infty .$
