Question:
Let \(f(x)\) be a differentiable function at \(x = a\) with \(f'(a) = 2\) and \(f(a) = 4\).
Then \(\displaystyle\lim_{x \to a} \frac{x f(a) - a f(x)}{x - a}\) equals :
Let \(f(x)\) be a differentiable function at \(x = a\) with \(f'(a) = 2\) and \(f(a) = 4\).
Then \(\displaystyle\lim_{x \to a} \frac{x f(a) - a f(x)}{x - a}\) equals :
Then \(\displaystyle\lim_{x \to a} \frac{x f(a) - a f(x)}{x - a}\) equals :
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The limit is of the form 0/0 at \(x=a\); apply L'Hospital's rule (or split the numerator) and use \(f'(a)=2\), \(f(a)=4\).
Updated On: Nov 4, 2026
- \(4 - 2a\)
- \(2a + 4\)
- \(a + 4\)
- \(2a - 4\)
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The Correct Option is A
Solution and Explanation
At \(x=a\), numerator \(= a f(a) - a f(a) = 0\) and denominator \(=0\), so the form is \(\frac{0}{0}\).
Apply L'Hospital's rule (differentiate numerator and denominator w.r.t. \(x\)):
\(\lim_{x\to a}\frac{f(a) - a f'(x)}{1} = f(a) - a f'(a)\)
\(= 4 - 2a\).
Answer: (1) \(4 - 2a\)
Apply L'Hospital's rule (differentiate numerator and denominator w.r.t. \(x\)):
\(\lim_{x\to a}\frac{f(a) - a f'(x)}{1} = f(a) - a f'(a)\)
\(= 4 - 2a\).
Answer: (1) \(4 - 2a\)
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