$\lim \sqrt{\dfrac{3 + (n^2 – 1)}{(3+n^2)} – \dfrac{1}{2^n}}=$ 27/10/2021 Bởi Eva $\lim \sqrt{\dfrac{3 + (n^2 – 1)}{(3+n^2)} – \dfrac{1}{2^n}}=$
$lim\sqrt[2]{\frac{3+(n^2-1)}{3+n^2}-\frac{1}{2^2}}$ $=lim\sqrt[2]{\frac{3+n^2}{3+n^2}-(\frac{1}{2})^n }$ $=lim\sqrt[2]{\frac{\frac{2}{n^2}+1}{\frac{2}{n^2}+1}-(\frac{1}{2})^n }$ $=\sqrt[2]{1}-0$ $=1$ Bình luận
Đáp án: Giải thích các bước giải: lim $\sqrt[]{\frac{3+(n^{2}-1)}{(3+ n^{2})}-\frac{1}{2^{n}}}$ =lim $\sqrt[]{\frac{2+n^{2}}{3+ n^{2}}-\frac{1^n}{2^n}}$ =lim $\sqrt[]{\frac{n^{2}(\frac{2}{n^2}+1)}{n^{2}(\frac{3}{n^2}+1)}-(\frac{1}{2})^{n}}$ =lim $\sqrt[]{\frac{\frac{2}{n^2}+1}{\frac{3}{n^2}+1}-(\frac{1}{2})^{n}}$ =$\sqrt[]{\frac{0+1}{0+1}-0}$ =1 Bình luận
$lim\sqrt[2]{\frac{3+(n^2-1)}{3+n^2}-\frac{1}{2^2}}$
$=lim\sqrt[2]{\frac{3+n^2}{3+n^2}-(\frac{1}{2})^n }$
$=lim\sqrt[2]{\frac{\frac{2}{n^2}+1}{\frac{2}{n^2}+1}-(\frac{1}{2})^n }$
$=\sqrt[2]{1}-0$
$=1$
Đáp án:
Giải thích các bước giải:
lim $\sqrt[]{\frac{3+(n^{2}-1)}{(3+ n^{2})}-\frac{1}{2^{n}}}$
=lim $\sqrt[]{\frac{2+n^{2}}{3+ n^{2}}-\frac{1^n}{2^n}}$
=lim $\sqrt[]{\frac{n^{2}(\frac{2}{n^2}+1)}{n^{2}(\frac{3}{n^2}+1)}-(\frac{1}{2})^{n}}$
=lim $\sqrt[]{\frac{\frac{2}{n^2}+1}{\frac{3}{n^2}+1}-(\frac{1}{2})^{n}}$
=$\sqrt[]{\frac{0+1}{0+1}-0}$ =1