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Left-hand limit at x = -1

Consider the piecewise function f(x)={x+3x+1x≤−1x2x>−1f(x) = \begin{cases} \dfrac{x+3}{x+1} & x \le -1 \\ x^2 & x > -1 \end{cases}

Evaluate lim⁡x→−1−f(x)\displaystyle\lim_{x \to -1^-} f(x).

Hint

For xx slightly less than −1-1, use the piece f(x)=x+3x+1f(x) = \dfrac{x+3}{x+1}. Think about what happens to the numerator and denominator as x→−1−x \to -1^-, and consider whether the expression can be simplified first.

Step-by-step solution

Since we are approaching −1-1 from the left, we use the rule f(x)=x+3x+1f(x) = \dfrac{x+3}{x+1} for x≤−1x \le -1.

As x→−1−x \to -1^-, we evaluate the behavior of this piece: the numerator x+3→2x+3 \to 2 and the denominator x+1→0−x+1 \to 0^- (approaching 0 from the negative side).

Since the numerator approaches a nonzero value (2) while the denominator approaches 0 from the negative side, the overall fraction approaches 20−=−∞\dfrac{2}{0^-} = -\infty.

Therefore, lim⁡x→−1−f(x)=−∞\displaystyle\lim_{x \to -1^-} f(x) = -\infty.

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