# What is the limit of (x - ln x) as x approaches oo?

Jan 22, 2017

This limit goes unbounded to $+ \infty$.

#### Explanation:

Trying to "substitute", gives us the indeterminate form $\infty - \infty$.

Recall that $x = \ln {e}^{x}$ and $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$:

${\lim}_{x \to \infty} \left(x - \ln x\right) = {\lim}_{x \to \infty} \left(\ln {e}^{x} - \ln x\right)$

$= {\lim}_{x \to \infty} \ln \left({e}^{x} / x\right)$. Let $u = {e}^{x} / x$.

Now we will take the limit of $u$ as $x$ approaches infinity:

${\lim}_{x \to \infty} {e}^{x} / x$. Since we get the indeterminate form $\frac{\infty}{\infty}$, and ${e}^{x}$ and $x$

are differentiable everywhere, we can use L'Hospital's rule:

${\lim}_{x \to \infty} \left({e}^{x} / x\right) = {\lim}_{x \to \infty} \frac{\left({e}^{x}\right) '}{\left(x\right) '}$

$= {\lim}_{x \to \infty} \left({e}^{x}\right) = \infty$.

Now we know that as $x \to \infty$, $u \to \infty$, and

${\lim}_{u \to \infty} \left(\ln u\right) = \infty$.

Therefore,

${\lim}_{u \to \infty} \left(\ln u\right) = {\lim}_{x \to \infty} \ln \left({e}^{x} / x\right)$

$= {\lim}_{x \to \infty} \left(\ln {e}^{x} - \ln x\right) = {\lim}_{x \to \infty} \left(x - \ln x\right)$.