Bài 2: Giới hạn của hàm số

\(lim\left(n-3-\sqrt{n^2-\sqrt{5}n+1}\right)=lim\dfrac{-6n+n\sqrt{5}+8}{n+3+\sqrt{n^2-\sqrt{5}n+1}}\)

=\(lim\dfrac{n\left(-6+\sqrt{5}+\dfrac{8}{n}\right)}{n\left(1+\dfrac{3}{n}+\sqrt{1-\dfrac{\sqrt{5}}{n}+\dfrac{1}{n^2}}\right)}=lim\dfrac{-6+\sqrt{5}+\dfrac{8}{n}}{1+\dfrac{3}{n}+\sqrt{1-\dfrac{\sqrt{5}}{n}+\dfrac{1}{n^2}}}=\dfrac{\sqrt{5}}{2}-3\)

\(\Rightarrow a=\dfrac{1}{2};b=-3\)\(\Rightarrow a+b=\dfrac{-5}{2}\)

\(=lim_{x\rightarrow1}\dfrac{-x^2+2x-1}{x\left(x-1\right)\left(\sqrt{2x-x^2}+1\right)}=lim_{x\rightarrow1}\dfrac{-\left(x-1\right)^2}{x\left(x-1\right)\left(\sqrt{2x-x^2}+1\right)}\)

\(=lim_{x\rightarrow1}-\dfrac{\left(x-1\right)}{x\left(\sqrt{2x-x^2}+1\right)}\)=0

\(\lim\limits_{ }\sqrt[3]{27n^3-7n^2}-3n=\lim\limits\dfrac{\left(27n^3-7n^2\right)-27n^3}{\left(\sqrt[3]{27n^3-7n^2}\right)^2+\sqrt[3]{27n^3-7n^2}.3n+9n^2}_{ }=\)

\(\lim\limits_{ }\dfrac{-7n^2}{n^2\left[\left(\sqrt[3]{27-\dfrac{7}{n}}\right)^2+\sqrt[3]{27-\dfrac{7}{n}}.3+9\right]}=\dfrac{-7}{27}\)

Nhờ các cao thủ giải dùm mình cái nha. Tks.

Tìm Lim ( )

bằng 1 bạn

a)lim\(\dfrac{4-3^n}{2.3^n+2}\)=lim\(\dfrac{4.\dfrac{1^n}{3^n}-\dfrac{3^n}{3^n}}{2.\dfrac{3^n}{3^n}+2\dfrac{1^n}{3^n}}=\)\(lim\dfrac{4.(\dfrac{1}{3})^n-1}{2.1+2.(\dfrac{1}{3})^n}=\dfrac{4.0-1}{\dfrac{2+2.0}{ }}=\dfrac{-1}{2}\)

b) lim\(\dfrac{3^{n+1}-2^n}{2-2.3^n}=lim\dfrac{3^n.3-2^n}{2-2.3^n}=lim\dfrac{3.\dfrac{3^n}{3^n}-\left(\dfrac{2}{3}\right)^n}{2.\left(\dfrac{1}{3}\right)^n-2.\dfrac{3^n}{3^n}}=\dfrac{3.1-0}{2.0-2.1}=\dfrac{-3}{2}\)

a)lấy (x^2-9) ra khử còn cái đó thay vô =-18/1=-18

a) lim= - 1/0 = - vô cùng

d) lim x(x^99-2)+1/ x(x^49-2)+1 =lim (x^99-2)/(x^49-2)=1