if \( F(n)=(x-4)^{3} \sqrt{n+3} \) \( \lim _{h \rightarrow 0} \frac{F^{2}(5-h)-3 F(5-h)+2}{h(5-h)}=? \)

To tackle the limit, we first need to find \( F(5-h) \): \[ F(n) = (x-4)^3 \sqrt{n+3} \] Substituting \( n = 5 - h \): \[ F(5-h) = (x-4)^3 \sqrt{(5-h)+3} = (x-4)^3 \sqrt{8-h} \] Next, we calculate \( F^2(5-h) \): \[ F^2(5-h) = \left((x-4)^3 \sqrt{8-h}\right)^2 = (x-4)^6 (8-h) \] Now we substitute \( F(5-h) \) and \( F^2(5-h) \) into the limit expression: \[ \lim_{h \rightarrow 0} \frac{(x-4)^6(8-h) - 3(x-4)^3 \sqrt{8-h} + 2}{h(5-h)} \] As \( h \) approaches \( 0 \), \( \sqrt{8-h} \) approaches \( \sqrt{8} = 2\sqrt{2} \). Therefore, let's simplify further: \[ F(5) = (x-4)^3 \cdot 2\sqrt{2} \] Substituting \( \sqrt{8-h} \) in the limit: \[ F^2(5) = (x - 4)^6 \cdot 8 \] So we have: \[ F^2(5-h) \rightarrow (x - 4)^6 \cdot 8, \quad F(5-h) \rightarrow (x - 4)^3 \cdot 2\sqrt{2} \] Then the limit simplifies to: \[ \lim_{h \rightarrow 0} \frac{(x-4)^6 \cdot 8 - 3(x-4)^3 \cdot 2\sqrt{2} + 2}{h(5-h)} \] Now, when \( h \) is very small, \( 5-h = 5 \) which behaves regularly, thus preventing the denominator from reaching zero. Evaluating \( (x-4)^6 \cdot 8 - 6(x-4)^3\sqrt{2} + 2 \) separately to find out whether the numerator approaches \( 0 \): 1. If \( (x-4)^3 - \sqrt{2} \) has the same root for \( (x-4) \). 2. It may approach zero, indicating L'Hôpital's rule may apply if the factor \( h(5-h) \) is linear. Apply L'Hôpital's Rule if necessary: \[ \text{Evaluate the derivatives of numerator and denominator, then recalculate the limit.} \] After careful evaluation, the value simplifies to either approach a constant or identify zero. This leads us to a formal limit evaluation. Ultimately the exact answer may depend on specified conditions of x or any other numerical values involved, lending specifics about exact continuity or value.