Giải bài 4.47 trang 172 - SBT Đại số và Giải tích lớp 11
Tính các giới hạn sau (\(n\to +\infty\))
a) \(\lim \dfrac{{{\left( -3 \right)}^{n}}+{{2.5}^{n}}}{1-{{5}^{n}}} \)
b) \(\lim \dfrac{1+2+3+...+n}{{{n}^{2}}+n+1} \)
c) \(\lim \left( \sqrt{{{n}^{2}}+2n+1}-\sqrt{{{n}^{2}}+n-1} \right) \)
a)
\( \lim \dfrac{{{\left( -3 \right)}^{n}}+{{2.5}^{n}}}{1-{{5}^{n}}} \\ =\lim \dfrac{{{\left( -\dfrac{3}{5} \right)}^{n}}+2}{{{\left( \dfrac{1}{5} \right)}^{n}}-1}=-2 \)
b)
\(\lim \dfrac{1+2+3+...+n}{{{n}^{2}}+n+1}\\=\lim \dfrac{\dfrac{n\left( n+1 \right)}{2}}{{{n}^{2}}+n+1} \\ =\lim \dfrac{{{n}^{2}}+n}{2\left( {{n}^{2}}+n+1 \right)} \\ =\lim \dfrac{1+\dfrac{1}{n}}{2\left( 1+\dfrac{1}{n}+\dfrac{1}{{{n}^{2}}} \right)}=\dfrac{1}{2} \)
c)
\(\lim \left( \sqrt{{{n}^{2}}+2n+1}-\sqrt{{{n}^{2}}+n-1} \right) \\ =\lim \dfrac{n+2}{\sqrt{{{n}^{2}}+2n+1}+\sqrt{{{n}^{2}}+n-1}} \\ =\lim \dfrac{1+\dfrac{2}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{1}{{{n}^{2}}}}+\sqrt{1+\dfrac{1}{n}-\dfrac{1}{{{n}^{2}}}}}=\dfrac{1}{2} \)