Question Number 98004 by Ar Brandon last updated on 11/Jun/20
![lim_(n→∞) [sin(n)+4^n ×(3/n^2 )×((n+1)/(n^2 −4))]](Q98004.png)
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{sin}\left(\mathrm{n}\right)+\mathrm{4}^{\mathrm{n}} ×\frac{\mathrm{3}}{\mathrm{n}^{\mathrm{2}} }×\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{4}}\right] \\ $$
Commented by bobhans last updated on 11/Jun/20
![lim_(n→∞) ((sin n)/n) × lim_(n→∞) ((3.4^n )/n) ×lim_(n→∞) ((n+1)/(n^2 −4)) = 0 × ∞ × 0 = 0 [ lim_(n→∞) ((3.4^n )/n) = 3 × lim_(n→∞) ((ln(4).4^n )/1) = ∞ ]](Q98013.png)
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{n}}{\mathrm{n}}\:×\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}.\mathrm{4}^{\mathrm{n}} }{\mathrm{n}}\:×\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{4}} \\ $$$$=\:\mathrm{0}\:×\:\infty\:×\:\mathrm{0}\:=\:\mathrm{0}\: \\ $$$$\left[\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}.\mathrm{4}^{\mathrm{n}} }{\mathrm{n}}\:=\:\mathrm{3}\:×\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{4}\right).\mathrm{4}^{\mathrm{n}} }{\mathrm{1}}\:=\:\infty\:\right]\: \\ $$