\(\underset{x\rightarrow 1}{lim}\) f(x) = \(\underset{x\rightarrow 1}{lim}\) \(\frac{x+x^2+x^3-3}{x-1}\) (0/0 type)
= \(\underset{x\rightarrow 1}{lim}\) \(\frac{x^3+x^2+x-3}{x-1}\)
= \(\underset{x\rightarrow 1}{lim}\) \(\frac{(x-1)(x^2+2x+3)}{x-1}\)
= \(\underset{x\rightarrow 1}{lim}\) (x2+2x+3)
= 12 + 2 x 1 + 3 (By taking limit)
= 1+2+3 = 6