Google
Câu hỏi: Tính giới hạn ${T=\lim \left( \sqrt{{{16}^{n+1}}+{{4}^{n}}}-\sqrt{{{16}^{n+1}}+{{3}^{n}}} \right)}$.
A. ${T=\dfrac{1}{16}}$.
B. ${T=0}$.
C. ${T=\dfrac{1}{4}}$.
D. ${T=\dfrac{1}{8}}$.
A. ${T=\dfrac{1}{16}}$.
B. ${T=0}$.
C. ${T=\dfrac{1}{4}}$.
D. ${T=\dfrac{1}{8}}$.
Ta có: $\sqrt{{{16}^{n+1}}+{{4}^{n}}}-\sqrt{{{16}^{n+1}}+{{3}^{n}}}=\dfrac{{{4}^{n}}-{{3}^{n}}}{\sqrt{{{16}^{n+1}}+{{4}^{n}}}+\sqrt{{{16}^{n+1}}+{{3}^{n}}}}=\dfrac{1-{{\left( \dfrac{3}{4} \right)}^{n}}}{\sqrt{4+\dfrac{1}{{{4}^{n}}}}+\sqrt{+{{\left( \dfrac{3}{16} \right)}^{n}}}}$
Vậy T $=\lim \left( \sqrt{{{16}^{n+1}}+{{4}^{n}}}-\sqrt{{{16}^{n+1}}+{{3}^{n}}} \right)=\lim \dfrac{1-{{\left( \dfrac{3}{4} \right)}^{n}}}{\sqrt{4+\dfrac{1}{{{4}^{n}}}}+\sqrt{+{{\left( \dfrac{3}{16} \right)}^{n}}}}=\dfrac{1}{4}$
Vậy T $=\lim \left( \sqrt{{{16}^{n+1}}+{{4}^{n}}}-\sqrt{{{16}^{n+1}}+{{3}^{n}}} \right)=\lim \dfrac{1-{{\left( \dfrac{3}{4} \right)}^{n}}}{\sqrt{4+\dfrac{1}{{{4}^{n}}}}+\sqrt{+{{\left( \dfrac{3}{16} \right)}^{n}}}}=\dfrac{1}{4}$
Đáp án C.