Question

Tìm các giới hạn sau:

a) \(lim\left(4^n-3^n\right)\)

b) \(lim\left[\left(2^n+1\right)^2-4^n\right]\)

c) \(lim\left(\sqrt{2n^5-3n^2+11}-n^3\right)\)

d) \(lim\left(\sqrt{2n^2+1}-\sqrt{3n^2-1}\right)\)

e) \(lim\sqrt{n^2+3n\sqrt{n}+1}-n\)

Answer 1

\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)

\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)

\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)

\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)