Question

f(x) is differentiable function and given f'(2)=6, f'(1)=4, then \(\lim_{h\rightarrow0}\frac{f(2+2h+h^2)-f(2)}{f(1+h-h^2)-f(1)}\)

• A. does not exist
• B. equal to -3
• C. equal to 3
• D. equal to 3/2

Answer 1

The Correct Option is C

Given:
\[ \lim_{h \to 0} \frac{f(1 + h - h^2) - f(1)}{f(2 + 2h + h^2) - f(2)} \] Since \( f(x) \) is differentiable, we can apply the first-order approximation: \[ f(a + \delta) \approx f(a) + f'(a) \cdot \delta \] Step 1: Numerator approximation
Let \( x = 1 + h - h^2 \). Then: \[ f(1 + h - h^2) \approx f(1) + f'(1) \cdot (h - h^2) \] \[ \Rightarrow f(1 + h - h^2) - f(1) \approx f'(1)(h - h^2) \] Step 2: Denominator approximation 
Let \( y = 2 + 2h + h^2 \). Then: \[ f(2 + 2h + h^2) \approx f(2) + f'(2) \cdot (2h + h^2) \] \[ \Rightarrow f(2 + 2h + h^2) - f(2) \approx f'(2)(2h + h^2) \] Step 3: Substitute in limit expression
\[ \lim_{h \to 0} \frac{f'(1)(h - h^2)}{f'(2)(2h + h^2)} \] Step 4: Plug in given values
Given: \( f'(1) = 4 \), \( f'(2) = 6 \) \[ = \lim_{h \to 0} \frac{4(h - h^2)}{6(2h + h^2)} = \frac{4}{6} \cdot \lim_{h \to 0} \frac{h - h^2}{2h + h^2} \] Step 5: Simplify the limit
\[ \lim_{h \to 0} \frac{h(1 - h)}{h(2 + h)} = \lim_{h \to 0} \frac{1 - h}{2 + h} = \frac{1}{2} \] Final result: \[ \frac{4}{6} \cdot \frac{1}{2} = \frac{2}{6} = \frac{1}{3} \] But this contradicts the earlier given answer as 3, so let's recheck using L’Hôpital's Rule:
Let: Numerator: \( N(h) = f(1 + h - h^2) - f(1) \) Denominator: \( D(h) = f(2 + 2h + h^2) - f(2) \) Apply Chain Rule:
\[ N'(h) = f'(1 + h - h^2)(1 - 2h) \] \[ D'(h) = f'(2 + 2h + h^2)(2 + 2h) \] Now take the limit:
\[ \lim_{h \to 0} \frac{f'(1 + h - h^2)(1 - 2h)}{f'(2 + 2h + h^2)(2 + 2h)} \Rightarrow \frac{f'(1)(1)}{f'(2)(2)} = \frac{4}{12} = \frac{1}{3} \] So, the correct answer is: Option (C): equal to \(\frac{1}{3}\) (The answer 3 is incorrect; the accurate value is \(\frac{1}{3}\))