Răspuns:
Explicație pas cu pas:
Separi in doua limite laterale, lim cand x<1 si lim cand x>1
In primul caz, modului se transforma in (1-x), in al doilea in (x-1)
In primul caz, fractia de sub limita devine:
[tex]\frac{(x+1)^3 - (1-x)^3-8}{x^2 - 3x + 2} =[/tex]
[tex]=\frac{(x+1)^3 + (x-1)^3-8}{(x-1)(x-2)} =[/tex]
[tex]=\frac{[(x+1)+(x-1)][(x+1)^2 -(x+1)(x-1) + (x-1)^2]-8}{(x-1)(x-2)} =[/tex]
[tex]=\frac{2x(x^2 + 2x + 1 -x^2+1+ x^2-2x+1)-8}{(x-1)(x-2)} =[/tex]
[tex]=\frac{2x(x^2 + 3)-8}{(x-1)(x-2)} =[/tex]
[tex]=\frac{2x^3 + 6x-8}{(x-1)(x-2)} =[/tex]
[tex]=\frac{2x^3 -2x^2 + 2x^2 -2x + 8x-8}{(x-1)(x-2)} =[/tex]
[tex]=\frac{(x-1)(2x^2 + 2x + 8)}{(x-1)(x-2)} =[/tex]
[tex]=\frac{2(x^2 + x + 4)}{x-2}[/tex]
care are limita -12
In al doilea caz, fractia de sub limita devine:
[tex]\frac{(x+1)^3 - (x-1)^3-8}{x^2 - 3x + 2} =[/tex]
[tex]=\frac{[(x+1)-(x-1)][(x+1)^2 +(x+1)(x-1) + (x-1)^2]-8}{(x-1)(x-2)} =[/tex]
[tex]=\frac{2(x^2 + 2x + 1 +x^2 - 1 + x^2-2x+1)-8}{(x-1)(x-2)} =\\[/tex]
[tex]=\frac{2(3x^2 + 1)-8}{(x-1)(x-2)} =[/tex]
[tex]=\frac{6x^2 -6}{(x-1)(x-2)} =[/tex]
[tex]=\frac{6(x^2-1)}{(x-1)(x-2)} =[/tex]
[tex]=\frac{6(x-1)(x+1)}{(x-1)(x-2)} =[/tex]
[tex]=\frac{6(x+1)}{x-2}[/tex]
care are limita -12
