Răspuns:
Explicație pas cu pas:
[tex]f(x)=\frac{1}{(x-1)^{2}}-\frac{1}{x^{2}} =(x-1)^{-2}-x^{-2} \\Deci~f '(x)=-2(x-1)^{-3}*(x-1)'-(-2)*x^{-3}*x'=\frac{-2}{(x-1)^{3}} +\frac{2}{x^{3}}=-2*\frac{x^{3}-(x-1)^{3}}{x^{3}(x-1)^{3}}=-2*\frac{(x-x+1)(x^{2}-x*(x-1)+(x-1)^{2}}{x^{3}(x-1)^{3}}=\frac{-2(x^{2}+x^{2}-x+x^{2}-2x+1)}{x^{3}(x-1)^{3}} \\Deci~~f '(x)=\frac{-2(3x^{2}-3x+1)}{x^{3}(x-1)^{3}} .\\[/tex]
b) Ecuatia tangentei, pentru x=2
y=f(2)+f '(2)·(x-2). Deoarece dreapta cautata este paralela tangentei, ⇒m=f '(2)=-2·(3·4-3·2+1)/(2³·1³)=-2·7/8=-7/4.
Deci, dreapta cautata are ecuatia g(x)=mx+b. Deoarece A(0;3)∈Gg ⇒g(0)=3, deci (-7/4)·0+b=3, ⇒b=3. Deci dreapta cautata este y=(-7/4)x+3.
[tex]c) ~f(2)=1-\frac{1}{2^{2}};~~f(3)=\frac{1}{2^{2}}-\frac{1}{3^{2}};~.... ;~f(n)=\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}.\\Deci~f(2)+f(3)+...+f(n)=1-\frac{1}{2^{2}}+\frac{1}{2^{2}}-\frac{1}{3^{2}}+...+\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}=1-\frac{1}{n^{2}}\\Deci~~ \lim_{n \to \infty} (f(2)+f(3)+...+f(n))^{n^{2}}= \lim_{n \to \infty} (1-\frac{1}{n^{2}})^{n^{2}}=???[/tex]
