1. The problem is to find the limit \( \lim_{x \to 2} \frac{x^2 - 8}{x - 2} \). 2. First, substitute \( x = 2 \) directly to check for an indeterminate form: \[ \frac{2^2 - 8}{2 - 2} = \frac{4 - 8}{0} = \frac{-4}{0} \] This is division by zero which is undefined, so we need to simplify the expression. 3. Factor the numerator \( x^2 - 8 \). Since 8 is not a perfect square, factorization as difference of squares is not possible, but we can try to express or rewrite the numerator differently. However, since direct substitution leads to division by zero and numerator \(x^2 - 8\) at \(x=2\) is -4, the limit may tend to infinity or negative infinity depending on the direction. 4. Examine the behavior as \( x \to 2^+ \) (values slightly greater than 2): - Numerator: \( x^2 - 8 \) is slightly greater than -4 (e.g., at 2.1, \(2.1^2 - 8 = 4.41 - 8 = -3.59\)) - Denominator: \( x - 2 \) is positive (0.1) - Fraction is negative divided by positive equals negative, approximately \( -3.59 / 0.1 = -35.9 \) 5. Examine the behavior as \( x \to 2^- \) (values slightly less than 2): - Numerator: slightly less than -4 (e.g., at 1.9, \(1.9^2 - 8 = 3.61 - 8 = -4.39\)) - Denominator: \( x - 2 \) is negative (-0.1) - Fraction is negative divided by negative equals positive, approximately \( -4.39 / -0.1 = 43.9 \) 6. Since the left-hand limit \( \lim_{x \to 2^-} \) tends to positive infinity and the right-hand limit \( \lim_{x \to 2^+} \) tends to negative infinity, the two-sided limit does not exist. Final answer: $$ \lim_{x \to 2} \frac{x^2 - 8}{x - 2} \text{ does not exist since the left and right limits differ.} $$