Expert Answer to "How would you find this limit ?limn→∞(1+ln⁡(1−12n))n"

Beryneingmk39

Answered question

2022-03-29

How would you find this limit ?

lim_{n \to \infty} {(1 + \ln (1 - \frac{1}{2 n}))}^{n}

sa3b4or9i9

Beginner2022-03-30Added 14 answers

We have

lim_{n \to \infty} {(1 + \ln (1 - \frac{1}{2 n}))}^{n}

= lim_{n \to \infty} ((1 + \ln (1 - \frac{1}{2 n} ({(^{{\frac{1}{\ln (1 - \frac{1}{2 n})}}})}^{n \ln (1 - \frac{1}{2 n})} = e^{lim_{n \to \infty} n \ln (1 - \frac{1}{2 n})}

provided the limit in the exponent exists. (Since

lim_{n \to \infty} ({(1 + \ln (1 - \frac{1}{2 n}))}^{\frac{1}{\ln (1 - \frac{1}{2 n})}}) = lim_{x \to 0} {(1 + x)}^{\frac{1}{x}} = e

)
So it only remains to compute the limit from the exponent. We get

lim_{n \to \infty} n \ln (1 - \frac{1}{2 n}) = - {\frac{12}{lim}}_{n \to \infty} \frac{\ln (1 - \frac{1}{2 n})}{- \frac{1}{2 n}} = - \frac{12}{}

using

lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1

So the original limit is

e^{- \frac{1}{2}}

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