Question

Valoarea limitei este...

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\displaystyle I_n=\int\limits_0^1n^3t^n(1-t)^2dt=\int\limits_1^0n^3t^n(1^2-2 \cdot 1 \cdot t+t^2)dt=\\ \\ =\int\limits_0^1n^3t^n(1-2t+t^2)dt=\int\limits_0^1(n^3t^n-2n^3t^nt+n^3t^nt^2)dt=\\ \\ =\int\limits_0^1(n^3t^n-2n^3t^{n+1}+n^3t^{n+2})dt=\int\limits_0^1n^3t^ndt-\int\limits_0^12n^3t^{n+1}dt+\int\limits_0^1n^3t^{n+2}dt\\ \\ =n^3\int\limits_0^1t^ndt-2n^3\int\limits_0^1t^{n+1}dt+n^3\int\limits_0^1t^{n+2}dt=$

$\displaystyle =n^3 \cdot \frac{t^{n+1}}{n+1} \Bigg|_0^1-2n^3 \cdot \frac{t^{n+2}}{n+2}\Bigg|_0^1+n^3 \cdot \frac{t^{n+3}}{n+3}\Bigg|_0^1=\\ \\ \\=n^3\left(\frac{1^{n+1}}{n+1}-\frac{0^{n+1}}{n+1} \right)-2n^3\left(\frac{1^{n+2}}{n+2}-\frac{n^{n+2}}{n+2} \right)+n^3\left(\frac{1^{n+3}}{n+3}+\frac{0^{n+3}}{n+3} \right)=\\ \\ \\=n^3 \cdot \frac{1}{n+1} -2n^3\cdot\frac{1}{n+2} +n^3\cdot \frac{1}{n+3} =\frac{n^3}{n+1} -\frac{2n^3}{n+2} +\frac{n^3}{n+3}=$

$\displaystyle =\frac{n^3(n+2)(n+3)-2n^3(n+1)(n+3)+n^3(n+1)(n+2)}{(n+1)(n+2)(n+3)} =\\ \\ \\=\frac{n^3(n^2+3n+2n+6)-2n^3(n^2+3n+n+3)+n^3(n^2+2n+n+2)}{(n+1)(n+2)(n+3)} =\\ \\ \\=\frac{n^3(n^2+5n+6)-2n^3(n^2+4n+3)+n^3(n^2+3n+2)}{(n+1)(n+2)(n+3)}=\\ \\\\ =\frac{n^5+5n^4+6n^3-2n^5-8n^4-6n^3+n^5+3n^4+2n^3}{(n+1)(n+2)(n+3)}=\\ \\ \\=\frac{2n^3}{(n+1)(n+2)(n+3)}$

$\displaystyle\lim_{n \to \infty}I_n= \lim_{n \to \infty}\frac{2n^3}{(n+1)(n+2)(n+3)} = \lim_{n \to \infty}\frac{2n^3}{n\left(\frac{1}{n} +1\right)n\left(\frac{2}{n} +1\right)n\left(\frac{3}{n}+1\right) }\\\\\\=\lim_{n \to \infty}\frac{2n^3}{n^3\left(\frac{1}{n}+1\right)\left(\frac{2}{n}+1\right)\left(\frac{3}{n}+1\right)}=\lim_{n\to\infty}\frac{2}{\left(\frac{1}{n}+1\right)\left(\frac{2}{n}+1\right)\left(\frac{3}{n}+1\right)}=\\ \\ \\=\frac{2}{(0+1)(0+1)(0+1)}=\frac{2}{1\cdot1\cdot1}=\frac{2}{1} =2$

$\displaystyle\mathbf{\lim_{n \to \infty}I_n=2~(b)}$