Question

Tìm các giới hạn sau:

a)\(lim\left[n^2\left(\sqrt{n^2+2}-\sqrt{n^2+4}\right)\right]\)

b)lim( \(\dfrac{3}{n-2}-5n\))

c) lim(\(\dfrac{n-1}{\sqrt{3}-n}-\dfrac{4}{2^{-n}}\))

d) \(lim\left(\dfrac{n^2-4}{n-2}-\dfrac{3n^2+4}{n}\right)\)

e) \(lim\dfrac{\sqrt{n^2+1}-n\sqrt{5}}{\sqrt{n^2+1}+n\sqrt{5}}\)

Answer 1

\(a=\lim\dfrac{-2n^2}{\sqrt{n^2+2}+\sqrt{n^2+4}}=\lim\dfrac{-2n}{\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1+\dfrac{4}{n^2}}}=\dfrac{-\infty}{2}=-\infty\)

\(b=\lim\dfrac{3-5n^2+10n}{n-2}=\lim\dfrac{-5n+10+\dfrac{3}{n}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)

\(c=\lim\left(\dfrac{1-\dfrac{1}{n}}{\dfrac{\sqrt{3}}{n}-1}-4.2^n\right)=-1-\infty=-\infty\)

\(d=\lim\dfrac{n^3-4n-\left(3n^2+4\right)\left(n-2\right)}{n^2-2n}=\lim\dfrac{-2n^3+6n^2-8n+8}{n^2-2n}\)

\(\lim\dfrac{-2n+6-\dfrac{8}{n}+\dfrac{8}{n^2}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)

\(e=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt{5}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{5}}=\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\)