lim-n-n-n-1-sin-1-n-n-where-Q-is-equal-to-

Question Number 63921 by raj last updated on 11/Jul/19

lim_(n→∞) (((n/(n+1)))^α +sin (1/n))^n where α∈Q is equal to

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{{n}}{{n}+\mathrm{1}}\right)^{\alpha} +\mathrm{sin}\:\frac{\mathrm{1}}{{n}}\right)^{{n}} \mathrm{where}\:\alpha\in\mathbb{Q} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Commented by kaivan.ahmadi last updated on 11/Jul/19

∼lim_(n→∞) (((n/(n+1)))^α −1+sin(1/n))(n)= lim_(n→∞) (((n^(α+1) −n(n+1)^α )/((n+1)^α ))+nsin(1/n))= ∼lim_(n→∞) (((−αn^α )/n^(α ) )+n×(1/n))=−α+1

$$\sim{lim}_{{n}\rightarrow\infty} \left(\left(\frac{{n}}{{n}+\mathrm{1}}\right)^{\alpha} −\mathrm{1}+{sin}\frac{\mathrm{1}}{{n}}\right)\left({n}\right)= \\ $$$${lim}_{{n}\rightarrow\infty} \left(\frac{{n}^{\alpha+\mathrm{1}} −{n}\left({n}+\mathrm{1}\right)^{\alpha} }{\left({n}+\mathrm{1}\right)^{\alpha} }+{nsin}\frac{\mathrm{1}}{{n}}\right)= \\ $$$$\sim{lim}_{{n}\rightarrow\infty} \left(\frac{−\alpha{n}^{\alpha} }{{n}^{\alpha\:} }+{n}×\frac{\mathrm{1}}{{n}}\right)=−\alpha+\mathrm{1} \\ $$

Commented by raj last updated on 11/Jul/19

$${thank}\:{you} \\ $$

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