Answer to Question #310402 in Real Analysis for Nikhil

Question #310402

For all even integral value of n, lim (x+1)^-n


n to ∞


Exist or not


True or false with full explanation



1
Expert's answer
2022-03-15T12:05:10-0400

limn(x+1)n\mathop {\lim }\limits_{n \to \infty } {\left( {x + 1} \right)^{ - n}}


=limnexp(ln((x+1)n))= \mathop {\lim }\limits_{n \to \infty } \exp \left( {\ln \left( {{{\left( {x + 1} \right)}^{ - n}}} \right)} \right)


=limnexp(nln(x+1))= \mathop {\lim }\limits_{n \to \infty } \exp \left( { - n\ln \left( {x + 1} \right)} \right)


=exp(ln(x+1)limnn)= \exp \left( { - \ln \left( {x + 1} \right) \cdot \mathop {\lim }\limits_{n \to \infty } n} \right)


Now for ln(x+1)>0\ln \left( {x + 1} \right) > 0


limn(x+1)n\mathop {\lim }\limits_{n \to \infty } {\left( {x + 1} \right)^{ - n}}


=exp(ln(x+1))= \exp \left( { - \ln \left( {x + 1} \right) \cdot \infty } \right)


=exp()= \exp \left( { - \infty } \right)


=0=0


Hence limn(x+1)n=0\mathop {\lim }\limits_{n \to \infty } {\left( {x + 1} \right)^{ - n}}=0 for ln(x+1)>0\ln \left( {x + 1} \right) > 0



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