\( \lim _ { x \rightarrow \pm 2 } \frac { 3 x + 4 } { x ^ { 2 } - 4 } \)

To evaluate the limit \[ \lim_{x \rightarrow \pm 2} \frac{3x + 4}{x^2 - 4}, \] we first observe the denominator \(x^2 - 4\). This can be factored as \[ x^2 - 4 = (x - 2)(x + 2). \] Next, we will evaluate the limit as \(x\) approaches \(2\) and \(-2\) separately. ### Limit as \(x \to 2\) Substituting \(x = 2\) into the denominator: \[ x^2 - 4 = 2^2 - 4 = 4 - 4 = 0. \] The numerator at \(x = 2\) is: \[ 3(2) + 4 = 6 + 4 = 10. \] Thus, we have a form of \(\frac{10}{0}\), which indicates that the limit may approach \(\infty\) or \(-\infty\). To determine the behavior, we check the left-hand limit and right-hand limit. #### Left-hand limit as \(x \to 2^-\): As \(x\) approaches \(2\) from the left, \(x - 2\) is negative and \(x + 2\) is positive. Therefore, the denominator \( (x - 2)(x + 2) \) is negative. The numerator \(3x + 4\) approaches \(10\) (positive). Thus, the limit is: \[ \lim_{x \to 2^-} \frac{3x + 4}{x^2 - 4} = \frac{10}{\text{negative}} = -\infty. \] #### Right-hand limit as \(x \to 2^+\): As \(x\) approaches \(2\) from the right, \(x - 2\) is positive and \(x + 2\) is still positive. Therefore, the denominator \( (x - 2)(x + 2) \) is positive. The numerator remains \(10\) (positive). Thus, the limit is: \[ \lim_{x \to 2^+} \frac{3x + 4}{x^2 - 4} = \frac{10}{\text{positive}} = +\infty. \] Since the left-hand limit approaches \(-\infty\) and the right-hand limit approaches \(+\infty\), we conclude that: \[ \lim_{x \to 2} \frac{3x + 4}{x^2 - 4} \text{ does not exist.} \] ### Limit as \(x \to -2\) Now we evaluate the limit as \(x\) approaches \(-2\). Substituting \(x = -2\) into the denominator: \[ x^2 - 4 = (-2)^2 - 4 = 4 - 4 = 0. \] The numerator at \(x = -2\) is: \[ 3(-2) + 4 = -6 + 4 = -2. \] Again, we have a form of \(\frac{-2}{0}\), indicating that the limit may approach \(\infty\) or \(-\infty\). We will check the left-hand limit and right-hand limit. #### Left-hand limit as \(x \to -2^-\): As \(x\) approaches \(-2\) from the left, \(x - 2\) is negative and \(x + 2\) is negative. Therefore, the denominator \( (x - 2)(x + 2) \) is positive. The numerator approaches \(-2\) (negative). Thus, the limit is: \[ \lim_{x \to -2^-} \frac{3x + 4}{x^2 - 4} = \frac{-2}{\text{positive}} = -\infty. \] #### Right-hand limit as \(x \to -2^+\): As \(x\) approaches \(-2\) from the right, \(x - 2\) is negative and \(x + 2\) is positive. Therefore, the denominator \( (x - 2)(x + 2) \) is negative. The numerator remains \(-2\) (negative). Thus, the limit is: \[ \lim_{x \to -2^+} \frac{3x + 4}{x^2 - 4} = \frac{-2}{\text{negative}} = +\infty.