1. **Stating the problem:** Find the limit \( \lim_{x \to 2} \frac{x^2 + 2x - 1}{x^2 - 4} \). 2. **Substitute \(x = 2\) directly:** Top: \(2^2 + 2 \times 2 - 1 = 4 + 4 - 1 = 7\) Bottom: \(2^2 - 4 = 4 - 4 = 0\) Since the denominator is zero, direct substitution leads to an undefined form. 3. **Factor the denominator:** \(x^2 - 4 = (x - 2)(x + 2)\) 4. **Check if numerator can be factored or simplified at \(x=2\):** Try to factor numerator \(x^2 + 2x - 1\) using quadratic formula for roots: Roots \(= \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}\) No root at \(x=2\), so numerator does not vanish at \(x=2\). 5. **Evaluate left and right limits:** Since denominator approaches zero but numerator approaches 7 (non-zero), the fraction approaches \(\pm\infty\) depending on side. For \(x \to 2^+\) (just greater than 2), denominator \((x-2)(x+2)\) is positive (since both factors \(> 0\)), so limit tends to \(+\infty\). For \(x \to 2^-\) (just less than 2), denominator \(x-2 < 0\) while \(x+2 > 0\), so product negative, thus limit tends to \(-\infty\). **Answer:** \( \lim_{x \to 2^+} \frac{x^2 + 2x -1}{x^2 -4} = +\infty, \quad \lim_{x \to 2^-} \frac{x^2 + 2x -1}{x^2 -4} = -\infty \) The overall limit does not exist because the left and right limits are not equal.