$\lim_{x \to \infty} (x - \ln x)$
When applying L'Hospital's Rule to
$$\lim_{x \to \infty} (x - \ln x)$$
I would have thought the answer to be
$$\lim_{x \to \infty} \left(1 - \frac{1}{x} \right) = 1$$
But the answers I am seeing are shown as
\begin{align} \lim_{x \to \infty} x \left(1 - \frac{\ln x}{x} \right) &= \lim_{x \to \infty} \left(\frac{x^2 - x \ln x}{x} \right) \\ &= \lim_{x \to \infty} \left(\frac{2x - 1 - \ln x}{1} \right) \\ &= \infty \end{align}
Any advice on why my approach might be wrong?
Thanks
$\endgroup$ 33 Answers
$\begingroup$If the L'Hopital formalism $\lim_{x\to c}{f(x)\over g(x)}=\lim_{x\to c}{f'(x)\over g'(x)}$ when ${f(x)\over g(x)}\to{\infty\over\infty}$ generalized to $\lim_{x\to c}(f(x)-g(x))=\lim_{x\to c}(f'(x)-g'(x))$ when $f(x)-g(x)\to\infty-\infty$, just because ${\infty\over\infty}$ and $\infty-\infty$ are both indeterminate, we would have obvious nonsense like
$$\lim_{x\to\infty}(3x-2x)=\lim_{x\to\infty}(3-2)=1$$
$\endgroup$ $\begingroup$Hint: $x- \ln x = \ln(e^x) + \ln(x^{-1}) = \ln (\frac{e^x}{x}).$
$\endgroup$ 1 $\begingroup$HINT
You can’t apply l’Hopital to the original limit which is in the form $\infty-\infty$, use that
$$x-\ln x=x\left(1-\frac{\ln x}{x}\right)$$
and apply l’Hopital to $$\lim_{x \to \infty}\frac{\ln x}{x}$$
which is in the form $\frac{\infty}{\infty}$.
$\endgroup$ 2
