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Evaluate the limits



Evaluate the limits ​-example-1

1 Answer

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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

User Robert McKee
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