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Evaluating an infinite limit near a vertical asymptote

Consider the function f(x)=x3+1(x+1)2f(x) = \dfrac{x^3+1}{(x+1)^2} near x=−1x = -1.

Determine lim⁡x→−1x3+1(x+1)2.\lim_{x \to -1} \frac{x^3+1}{(x+1)^2}. Factor the numerator first to simplify the expression, then analyze the behavior of the simplified function as x→−1x \to -1. State the value of the limit (using +∞+\infty or −∞-\infty if appropriate), and briefly justify whether the limit exists in the finite sense, and whether x=−1x=-1 gives a vertical asymptote of ff.

Hint

Factor x3+1x^3+1 as a sum of cubes: x3+1=(x+1)(x2−x+1)x^3+1 = (x+1)(x^2-x+1). Cancel the common factor of (x+1)(x+1) with part of the denominator, then examine the sign and size of what remains as x→−1x \to -1 from both sides.

Step-by-step solution

Factor the numerator using the sum of cubes formula: x3+1=(x+1)(x2−x+1)x^3+1 = (x+1)(x^2-x+1). This lets us rewrite f(x)=(x+1)(x2−x+1)(x+1)2=x2−x+1x+1f(x) = \frac{(x+1)(x^2-x+1)}{(x+1)^2} = \frac{x^2-x+1}{x+1} for x≠−1x \neq -1, canceling one factor of (x+1)(x+1).

Now evaluate the behavior as x→−1x \to -1: the numerator x2−x+1→1+1+1=3x^2-x+1 \to 1+1+1 = 3, a positive constant, while the denominator x+1→0x+1 \to 0. Since the denominator changes sign around x=−1x=-1 (negative for xx slightly less than −1-1, positive for xx slightly greater), the one-sided limits are −∞-\infty and +∞+\infty respectively.

Since the two one-sided limits disagree in sign, the two-sided limit does not exist as a finite number or as a single signed infinity; we say lim⁡x→−1f(x)\lim_{x\to -1} f(x) does not exist (it is −∞-\infty from the left and +∞+\infty from the right). Because at least one one-sided limit is infinite and ff is undefined at x=−1x=-1, the line x=−1x=-1 is a vertical asymptote of ff.

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