To compute lim n ( 1 + n ) 1 ln ⁡ n without L'Hospital

Question

To compute       lim    n    (  1  +  n      )                  1                  ln          ⁡          n                     without L&#039;Hospital

Aaron Everett · Accepted Answer

I always treat limits of this form by computing the limit of the logarithm, so we want to consider

lim_{n \to \infty} \frac{\ln (1 + n)}{\ln n}

If we consider instead of the sequence the function limit

lim_{x \to \infty} \frac{\ln (1 + x)}{\ln x}

we know that if this one exists, then it's the same of the limit of the sequence (the converse is generally not true). Now we can perform the substitution

x = 1 / t

and note that

\ln (1 + 1 / t) = \ln (t + 1) - \ln t

so the limit becomes

lim_{t \to 0^{+}} \frac{\ln t - \ln (1 + t)}{\ln t} = lim_{t \to 0^{+}} (1 - \frac{\ln (1 + t)}{\ln t}) = 1

because the fraction is

0 / \infty

Thus we have proved that

lim_{n \to \infty} (1 + n)^{1 / \ln n} = e^{1} = e

To compute <munder> <mo movablelimits="true" form="prefix">lim n </munder> ( 1 +

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