lim\(\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}\right)\)

= lim \(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}\right)\)

= lim \(\left(\frac{1}{3}-\frac{1}{n+1}\right)\)

= 1/3

\(\text{GIẢI :}\)

\(lim\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{\text{n}\left(\text{n}+1\right)}\right)\)

\(=lim\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{\text{n}}-\frac{1}{\text{n}+1}\right)\)

\(=lim\left(\frac{1}{3}-\frac{1}{\text{n}\left(\text{n + 1}\right)}\right)\)

\(=\frac{1}{3}\)

lim ( x ----> 0 ) \(\frac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{x}\)

= lim ( x----> 0 ) \(\frac{\sqrt[m]{1+ax}-1+1-\sqrt[n]{1+bx}}{x}\)

= lim ( x ---> 0 ) \(\frac{\sqrt[m]{1+ax}-1}{x}\)- lim ( x ---> 0 ) \(\frac{\sqrt[n]{1+bx}-1}{x}\)

= lim ( x ----> 0 ) \(\frac{ax}{x\left(\sqrt[m]{\left(1+ax\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1\right)}\)

- lim ( x ----> 0 ) \(\frac{bx}{x\left(\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1\right)}\)

= lim ( x -----> 0 ) \(\frac{a}{\sqrt[m]{\left(1+ax\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)

- lim ( x ---> 0 )  \(\frac{b}{\sqrt[n]{\left(1+bx\right)^{n-1}}+\sqrt[n]{\left(1+bx\right)^{n-2}}+...+1}\)

= \(\frac{a}{m}-\frac{b}{n}\)

cảm ơn bạn

woa ai ni

\(lim\frac{\sqrt{n^6-n+1}+n^2}{3n^2\sqrt{n^2-1}}=lim\frac{\sqrt{n^6-n+1}+n^2}{3\sqrt{n^6-n^4}}\)

\(=lim\frac{\sqrt{1-\frac{1}{n^5}+\frac{1}{n^6}}+\frac{1}{n}}{3\sqrt{1-\frac{1}{n^2}}}\)(chia cả tử và mẫu cho n³

\(=\frac{\sqrt{1-0+0}+0}{3\sqrt{1-0}}=\frac{1}{3}\)

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

\(=lim\left|n\right|\left(\sqrt{2+\frac{1}{n^2}}+\sqrt{2-\frac{1}{n^2}}\right)\)

Do \(lim\left|n\right|=+\infty\)

\(lim\left(\sqrt{2+\frac{1}{n^2}}+\sqrt{2-\frac{1}{n^2}}\right)=2\sqrt{2}\)

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

\(lim\frac{1+2\cdot3^n-7^n}{5^n+2\cdot7^n}\)

\(=lim\frac{\frac{1}{7^n}+\frac{6^n}{7^n}-1}{\frac{5^n}{7^n}+\frac{14^n}{7^n}}\)

\(=lim\frac{0+\left(\frac{6}{7}\right)^n-1}{\left(\frac{5}{7}\right)^n+2}=\frac{-1}{2}\)

\(lim\frac{\sqrt{4n^2+1}+2n-1}{\sqrt{n^2+4n+1}+n}\)

= \(lim\frac{\sqrt{4+\frac{1}{n^2}}+2-\frac{1}{n}}{\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+1}\)

=\(\frac{2+2}{1+1}=2\)