Question:
If \(f(x)\) is a differentiable function of \(x\), then \(\lim _{h\rightarrow 0}\frac{[f(x+h)]^{2}-[f(x)]^{2}}{2h}=\)
If \(f(x)\) is a differentiable function of \(x\), then \(\lim _{h\rightarrow 0}\frac{[f(x+h)]^{2}-[f(x)]^{2}}{2h}=\)
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Factorise the numerator as a difference of squares and use the definition of the derivative.
Updated On: Nov 12, 2026
- \([f(x)]^{2}\)
- \(f(x)\cdot f'(x)\)
- \(f(x)+f'(x)\)
- \([f(x)\cdot f'(x)]^{2}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The numerator is a difference of squares, \(a^{2}-b^{2}=(a-b)(a+b)\).
Step 2: Factorise
\([f(x+h)]^{2}-[f(x)]^{2}=[f(x+h)-f(x)]\,[f(x+h)+f(x)]\).
Step 3: Take the limit
\(\lim_{h\rightarrow0}\dfrac{f(x+h)-f(x)}{h}=f'(x)\) and \(f(x+h)+f(x)\rightarrow2f(x)\).
Step 4: Result
The limit is \(\dfrac{f'(x)\cdot2f(x)}{2}=f(x)\,f'(x)\).
Final Answer:
The limit equals f(x) times f'(x). \[ \boxed{\text{(B) }f(x)\cdot f'(x)} \]
The numerator is a difference of squares, \(a^{2}-b^{2}=(a-b)(a+b)\).
Step 2: Factorise
\([f(x+h)]^{2}-[f(x)]^{2}=[f(x+h)-f(x)]\,[f(x+h)+f(x)]\).
Step 3: Take the limit
\(\lim_{h\rightarrow0}\dfrac{f(x+h)-f(x)}{h}=f'(x)\) and \(f(x+h)+f(x)\rightarrow2f(x)\).
Step 4: Result
The limit is \(\dfrac{f'(x)\cdot2f(x)}{2}=f(x)\,f'(x)\).
Final Answer:
The limit equals f(x) times f'(x). \[ \boxed{\text{(B) }f(x)\cdot f'(x)} \]
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