Explain how to find the value of each limit 5) \( \lim _{x \rightarrow-2} \frac{(x+5)^{3}(x-1)^{2}}{(x+2)^{3}(x-2)^{2}} \)

We want to evaluate   limₓ→₋₂ [(x + 5)³ (x – 1)²] / [(x + 2)³ (x – 2)²]. Step 1. Look for any common factors.   Write down the factors in the numerator and denominator:   • Numerator: (x + 5)³ and (x – 1)².   • Denominator: (x + 2)³ and (x – 2)².   Notice that there is no (x + 2) factor in the numerator to cancel the (x + 2)³ in the denominator. Step 2. Check what happens when you plug in x = –2 (if possible).   • For the numerator:    – (x + 5) becomes (–2 + 5) = 3, so (x + 5)³ = 3³ = 27.    – (x – 1) becomes (–2 – 1) = –3, so (x – 1)² = (–3)² = 9.    Thus, the numerator → 27 × 9 = 243.   • For the denominator:    – (x + 2) becomes (–2 + 2) = 0, so (x + 2)³ → 0.    – (x – 2) becomes (–2 – 2) = –4, so (x – 2)² = (–4)² = 16.    Thus, the denominator → 0 × 16 = 0. Step 3. Interpret the behavior.   Since the numerator tends to 243 (a nonzero constant) and the denominator tends to 0, the fraction is “blowing up.” However, the sign of the limit depends on the behavior of the factor (x + 2)³ since it is the only one that becomes 0 and its exponent is odd.   • For x approaching –2 from the right (x → –2⁺):    – x + 2 is a small positive number,    – Therefore, (x + 2)³ is positive (since a small positive number cubed is positive).    – The numerator is positive (243), and the other factors in the denominator (like (x – 2)²) are positive.    – So overall, the expression tends to +∞.   • For x approaching –2 from the left (x → –2⁻):    – x + 2 is a small negative number,    – Since the exponent is 3 (an odd power), (x + 2)³ is negative.    – The numerator is still positive and (x – 2)² is positive,    – Hence, the overall expression tends to –∞. Step 4. Conclude the two-sided limit.   Because the limit from the right is +∞ and from the left is –∞, the two one-sided limits do not agree. Therefore, the two-sided limit as x → –2 does not exist. Summary:   The limit does not exist. Specifically, as x → –2⁺ the function tends to +∞ and as x → –2⁻ it tends to –∞.