How would you find this limit ? \(\displaystyle\lim_{{{n}\to\infty}}{\left({1}+{\ln{{\left({1}-{\frac{{{1}}}{{{2}{n}}}}\right)}}}\right)}^{{n}}\)

Beryneingmk39 2022-03-29 Answered
How would you find this limit ?
limn(1+ln(112n))n
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Answers (1)

sa3b4or9i9
Answered 2022-03-30 Author has 14 answers
We have
limn(1+ln(112n))n
=limn((1+ln(112n(({1ln(112n)})nln(112n)=elimnnln(112n)
provided the limit in the exponent exists. (Since
limn((1+ln(112n))1ln(112n))=limx0(1+x)1x=e )
So it only remains to compute the limit from the exponent. We get
limnnln(112n)=12limnln(112n)12n=12
using limx0ln(1+x)x=1
So the original limit is e12.
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