a)
[tex]f(x) = \dfrac{(x+3)(x-1)^2}{x^2} \\ \\ \lim\limits_{x\to \infty}\Big(\dfrac{(x+3)(x-1)^2}{x^2}-mx-n\Big) = 0 \\\\\lim\limits_{x\to \infty}\dfrac{(x+3)(x-1)^2-mx^3-nx^2}{x^2}= 0\\ \\ \lim\limits_{x\to \infty}\dfrac{x^3+x^2-5x+3-mx^3-nx^2}{x^2}= 0\\ \\ \lim\limits_{x\to \infty}\dfrac{x^3(1-m)+x^2(1-n)-5x+3}{x^2}= 0\\ \\ \Rightarrow 1-m = 0\text{ si }1-n = 0\Rightarrow m = 1\text{ si }n = 1 \\ \\ \Rightarrow y = mx+n \Rightarrow\boxed{y = x+1}\to \text{asimptota oblica spre }+\infty[/tex]
b)
Egalezi (x-1)(x²+x+6) = 0 => x-1 = 0 sau x²+x+6 = 0 (Δ < 0) =>
=> x = 1 sau x ∈ ∅ => x = 1 punct de extrem local al functiei.
