Question

Evaluate the limits​

Answer 1

$x > \ln(x)$ for all $x$, so

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}$

Similarly, $\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}$

We can of course see the limits are identical by replacing $x\mapsto e^x$, so that

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)$

You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

$\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)$

Using the rule, the limit here is

$\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0$

so the overall limit is

$\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right) = \ln(0) = \lim_{x\to0^+} \ln(x) = -\infty$