Question

If \[ f(x) = 3x^{15} - 5x^{10} + 7x^5 + 50\cos(x - 1), \] then \[ \lim_{h \to 0} \frac{f(1 - h) - f(1)}{h^2 + 3h} \] is:

• A. \( -25 \)
• B. \( 25 \)
• C. \( -10 \)
• D. \( 10 \)

Hint:

For limits involving function differences, use Taylor expansions to approximate terms and simplify calculations.

Answer 1

The Correct Option is C

Step 1: Compute \( f(1) \) 
Substituting \( x = 1 \) in \( f(x) \): \[ f(1) = 3(1)^{15} - 5(1)^{10} + 7(1)^5 + 50\cos(1 - 1). \] \[ = 3(1) - 5(1) + 7(1) + 50\cos(0). \] \[ = 3 - 5 + 7 + 50(1). \] \[ = 55. \] 

Step 2: Compute \( f(1 - h) \) using Taylor Expansion 
Expanding \( f(1 - h) \) using first-order approximations: \[ f(1 - h) = 3(1 - h)^{15} - 5(1 - h)^{10} + 7(1 - h)^5 + 50\cos(1 - h - 1). \] Approximating terms using \( (1 - h)^n \approx 1 - nh \) for small \( h \): \[ (1 - h)^{15} \approx 1 - 15h, \quad (1 - h)^{10} \approx 1 - 10h, \quad (1 - h)^5 \approx 1 - 5h. \] For cosine, \[ \cos(-h) \approx 1 - \frac{h^2}{2}. \] Substituting these approximations: \[ f(1 - h) \approx 3(1 - 15h) - 5(1 - 10h) + 7(1 - 5h) + 50(1 - \frac{h^2}{2}). \] \[ = (3 - 45h) - (5 - 50h) + (7 - 35h) + (50 - 25h^2). \] \[ = 3 - 5 + 7 + 50 - 45h + 50h - 35h - 25h^2. \] \[ = 55 - 30h - 25h^2. \] 

Step 3: Compute the difference \( f(1 - h) - f(1) \) 
\[ f(1 - h) - f(1) = (55 - 30h - 25h^2) - 55. \] \[ = -30h - 25h^2. \] 

Step 4: Compute the limit 
\[ \lim_{h \to 0} \frac{f(1 - h) - f(1)}{h^2 + 3h}. \] Substituting: \[ \lim_{h \to 0} \frac{-30h - 25h^2}{h^2 + 3h}. \] Factor \( h \) from the numerator and denominator: \[ \lim_{h \to 0} \frac{h(-30 - 25h)}{h(h + 3)}. \] Cancel \( h \): \[ \lim_{h \to 0} \frac{-30 - 25h}{h + 3}. \] Substituting \( h = 0 \): \[ \frac{-30}{3} = -10. \]

 Step 5: Conclusion 
Thus, the correct answer is: \[ \mathbf{-10}. \]