Question:
If \(f(1)=2,\ f'(1)=1\), then \(\lim_{x \to 1} \frac{x f(1) - f(x)}{x-1}\) is
If \(f(1)=2,\ f'(1)=1\), then \(\lim_{x \to 1} \frac{x f(1) - f(x)}{x-1}\) is
Show Hint
Try rewriting expressions to match derivative definition.
Updated On: Apr 30, 2026
- $0$
- $-2$
- $-1$
- $2$
- $1$
Show Solution
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The Correct Option is
Solution and Explanation
Concept:
Use derivative definition:
\[
f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}
\]
Step 1: Rewrite numerator
\[ x f(1) - f(x) = 2x - f(x) \]
Step 2: Split expression
\[ \frac{2x - f(x)}{x-1} = \frac{2x-2 + 2 - f(x)}{x-1} \] \[ = \frac{2(x-1)}{x-1} + \frac{2 - f(x)}{x-1} \] \[ = 2 - \frac{f(x)-2}{x-1} \]
Step 3: Apply limit
\[ = 2 - f'(1) = 2 - 1 = 1 \] Final Conclusion:
Option (E)
Step 1: Rewrite numerator
\[ x f(1) - f(x) = 2x - f(x) \]
Step 2: Split expression
\[ \frac{2x - f(x)}{x-1} = \frac{2x-2 + 2 - f(x)}{x-1} \] \[ = \frac{2(x-1)}{x-1} + \frac{2 - f(x)}{x-1} \] \[ = 2 - \frac{f(x)-2}{x-1} \]
Step 3: Apply limit
\[ = 2 - f'(1) = 2 - 1 = 1 \] Final Conclusion:
Option (E)
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