Question

Evaluate the limits

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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) (e^x)/(e^(e^x))\right)`

Using the rule, the limit here is

`\displaystyle \lim_(x\to\infty) ((e^x)')/(\left(e^(e^x)\right)') = \lim_(x\to\infty) (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) (e^x)/(e^(e^x))\right) = \ln(0) = \lim_(x\to0^+) \ln(x) = -\infty`