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limf(x) =x+x2 +x3 -3/x-1 where x tends to 1 |
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Answer» \(\underset{x\rightarrow 1}{lim}\) 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 |
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