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One-sided infinite limit as x approaches 2 from the left

Consider the function h(x)=x+1x2−4h(x) = \dfrac{x+1}{x^2-4}. Note that x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2), so hh has vertical asymptotes at x=2x=2 and x=−2x=-2.

Evaluate lim⁡x→2−x+1x2−4.\lim_{x \to 2^-} \frac{x+1}{x^2-4}. If the limit is infinite, state whether it is +∞+\infty or −∞-\infty.

Hint

Factor the denominator as (x−2)(x+2)(x-2)(x+2). Plug in a number slightly less than 2 (like x=1.9x=1.9) to check the sign of the numerator and each factor in the denominator, then determine the overall sign of the quotient.

Step-by-step solution

As x→2−x \to 2^-, the numerator x+1→3x+1 \to 3, a positive number. The denominator factors as (x−2)(x+2)(x-2)(x+2), and as x→2−x \to 2^- we have x−2→0−x-2 \to 0^- (a small negative number) while x+2→4x+2 \to 4 (positive). So the denominator behaves like a small negative number times a positive number, giving a small negative number.

Dividing a positive numerator by a denominator that is a small negative number approaching 0 produces values that grow without bound in the negative direction. Therefore lim⁡x→2−x+1x2−4=−∞\lim_{x \to 2^-} \frac{x+1}{x^2-4} = -\infty.

One-sided infinite limit as x approaches -2 from the right

Evaluate lim⁡x→−2+x+1x2−4.\lim_{x \to -2^+} \frac{x+1}{x^2-4}. If the limit is infinite, state whether it is +∞+\infty or −∞-\infty.

Hint

Factor the denominator as (x−2)(x+2)(x-2)(x+2). Consider a value slightly greater than −2-2 (like x=−1.9x=-1.9) to determine the sign of the numerator and each factor of the denominator, then find the sign of the whole expression.

Step-by-step solution

As x→−2+x \to -2^+, the numerator x+1→−1x+1 \to -1, a negative number. The denominator factors as (x−2)(x+2)(x-2)(x+2), and as x→−2+x \to -2^+ we have x+2→0+x+2 \to 0^+ (a small positive number) while x−2→−4x-2 \to -4 (negative). So the denominator behaves like a small positive number times a negative number, giving a small negative number.

Dividing a negative numerator by a denominator that is a small negative number approaching 0 produces values that grow without bound in the positive direction. Therefore lim⁡x→−2+x+1x2−4=+∞\lim_{x \to -2^+} \frac{x+1}{x^2-4} = +\infty.

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