Question

ajutor , dau 50 de puncte am mare nevoie​

Answer 1

a)

$\displaystyle\lim_{x\to0}\dfrac{x^3-2x^2}{3x^2+x}=\lim_{x\to0}\dfrac{x(x^2-2x)}{x(3x+1)}=\lim_{x\to0}\dfrac{x^2-2x}{3x+1}=\frac{0^2-2\cdot0}{3\cdot0+1}=\frac{0}{1}=0$

b)

$\displaystyle\lim_{x\to2}\frac{x^2-2x}{x^2-4}=\lim_{x\to2}\frac{x(x-2)}{(x-2)(x+2)}=\lim_{x\to2}\frac{x}{x+2}=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}$

c)

$\displaystyle\lim_{x\to1}\frac{x^2+3x-4}{x-1}=\lim_{x\to1}\frac{x^2+4x-x-4}{x-1}=\lim_{x\to1}\frac{x(x+4)-(x+4)}{x-1}=\\\lim_{x\to1}\frac{(x+4)(x-1)}{x-1}=\lim_{x\to1}(x+4)=1+4=5$