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Exercise: Limits of Polynomial and Ra... · Q16

Q.Evaluate lim⁡x→−1x3+1x+1\lim_{x\to-1}\dfrac{x^3+1}{x+1}.

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Direct substitution gives 00\dfrac{0}{0}, so factor using the sum-of-cubes identity x3+1=(x+1)(x2−x+1)x^3+1=(x+1)(x^2-x+1):

x3+1x+1=(x+1)(x2−x+1)x+1=x2−x+1(x≠−1).\frac{x^3+1}{x+1} = \frac{(x+1)(x^2-x+1)}{x+1} = x^2-x+1 \quad (x\neq-1). …

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