Tìm lim (n^2 - 2^n)

L

lovelycat_handoi95

[TEX]\blue{Voi\ \forall n \in N*\ ta\ co\ : n^2 \leq 2^n \\ \Rightarrow 0 < \frac{n^2}{2^n} \leq 1 \\ Ta\ co\ :\\ lim n^2 - 2^n \\ = {\frac{n^4-2^{2n}}{n^2+2^n}}\\ =lim {\frac{\frac{n^4}{2^{2n}}-1}{\frac{n^2}{2^{2n}}+\frac{2^n}{2^{2n}}}}\\ = lim {\frac{(\frac{n^2}{2^{n}})^2-1}{(\frac{n}{2^{n}})^2+(\frac{2}{2^{2}})^n}}\\ Lai\ co\ :\\\left{lim(\frac{n^2}{2^{n}})^2-1= -1 <0 \\ lim (\frac{n}{2^{n}})^2+(\frac{2}{2^{2}})^n=0 \\ (\frac{n}{2^{n}})^2+(\frac{2}{2^{2}})^n>0[/TEX]


[TEX]\blue{\Rightarrow lim {\frac{(\frac{n^2}{2^{n}})^2-1}{(\frac{n}{2^{n}})^2+(\frac{2}{2^{2}})^n}} = - \infty \\ \Rightarrow lim n^2 - 2^n = - \infty [/TEX]
 
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