Question

Dacă știe careva sa rezolve punctele b). și c). Va rog.​

Answer 1

$b)\: A(0, 3)\in d\\d||t\Leftrightarrow m_d=m_t\\\text{Tangenta \^in }x=2\Rightarrow t:y-f(2)=f'(2)(x-2)\\\displaystyle f(2)=\frac{1}{(2-1)^2}-\frac14=\frac34\\\\f'(2)=\frac{-2(3\cdot4-3\cdot2+1)}{2^3(2-1)^3}=-\frac74\\\\t:y-\frac34=-\frac74(x-2)\Rightarrow y=-\frac74x+\frac{17}4\Rightarrow m_t=-\frac74\\\Rightarrow m_d=-\frac74\\d:y-y_A=m_d(x-x_A)\Rightarrow y-3=-\frac74(x-0)\\d: y=-\frac74x+3$

$c)\displaystyle\lim\limits_{n\to\infty}\left(f(2)+f(3)+\dots+f(n)\right)^{n^2}=\\=\lim\limits_{n\to\infty}\left(1-\frac1{2^2}+\frac1{2^2}-\frac1{3^2}+\dots+\frac1{(n-1)^2}-\frac1{n^2}\right)^{n^2}=\\=\lim\limits_{n\to\infty}\left(1-\frac1{n^2}\right)^{n^2}=\lim\limits_{n\to\infty}\left(1+\frac{-1}{n^2}\right)^{n^2}\overset{[1^\infty]}{=}\\\\=\lim\limits_{n\to\infty}\left[\left(1+\frac{-1}{n^2}\right)^{-n^2}\right]^{\frac{-1}{n^2}\cdot n^2}=e^{\lim\limits_{n\to\infty}\frac{-1}{n^2}\cdot n^2}=e^{-1}$