Question

Ex 2 subpunctele b si c...

Answer 1

$\boxed{Acesta \:\ este \:\ punctul \:\ C}$

[tex]\int^n_0 \frac{dx}{(x+1)(x+2)} = \int \frac{1}{\left(x+1\right)\left(x+2\right)}dx =\int \frac{1}{x+1}-\frac{1}{x+2}dx =\int \frac{1}{x+1}dx-\int \frac{1}{x+2}dx \\ \\ \\Schimbare \:\ de \:\ variabila: =\ \textgreater \ \ln \left|x+1\right|-\ln \left|x+2\right| \\ \\ \\=\ \textgreater \ \frac{1}{2}\ln \left(\frac{\left(n+1\right)^2}{\left(n+2\right)^2}\right)-\left(-\ln \left(2\right)\right) =\ \textgreater \ \boxed{simplificare} = \ln \left(\frac{n+1}{n+2}\right)+\ln \left(2\right) [/tex]

[tex]\\ =\ \textgreater \ \lim_{\rightarrow\infty} n[ln2-\ln \left(\frac{n+1}{n+2}\right)+\ln \left(2\right)] \\ =\ \textgreater \ \lim _{n\to a}\left[f\left(n\right)\cdot g\left(n\right)\right]=\lim _{n\to a}f\left(n\right)\cdot \lim _{n\to a}g\left(n\right) \\=\ \textgreater \ \lim _{n\to \infty \:}\left(n\right)\cdot \lim \:_{n\to \infty \:}\left(\ln \left(2\right)-\ln \left(\frac{n+1}{n+2}\right)+\ln \left(2\right)\right) \\=\ \textgreater \ \lim _{n\to \infty \:}\left(n\right)=\infty \: [/tex]

[tex]\\ si \:\ \lim _{n\to \infty \:}\left(\ln \left(2\right)-\ln \left(\frac{n+1}{n+2}\right)+\ln \left(2\right)\right) \\= \lim _{n\to \infty \:}\left(\ln \left(2\right)\right)-\lim _{n\to \infty \:}\left(\ln \left(\frac{n+1}{n+2}\right)\right)+\lim _{n\to \infty \:}\left(\ln \left(2\right)\right) \\ \lim _{n\to \infty \:}\left(\ln \left(\frac{n+1}{n+2}\right)\right) =\ \textgreater \ g\left(n\right)=\frac{n+1}{n+2},\:f\left(u\right)=\ln \left(u\right) [/tex]

[tex]\lim _{n\to \infty \:}\left(\frac{n+1}{n+2}\right) =\lim _{n\to \infty \:}\left(\frac{1+\frac{1}{n}}{1+\frac{2}{n}}\right) =\frac{\lim _{n\to \infty \:}\left(1+\frac{1}{n}\right)}{\lim _{n\to \infty \:}\left(1+\frac{2}{n}\right)} = \frac{1}{1} = 1 \\ \lim _{u\to \:1}\left(\ln \left(u\right)\right) =\ln \left(1\right) = 0 \\ Deci \:\ rezultatul \:\ final \:\ este \:\ =\ln \left(2\right)-0+\ln \left(2\right) \\ =2\ln \left(2\right) =\infty \cdot \:2\ln \left(2\right) = \boxed{\infty \: }[/tex]