Discussion
If \(f(1) = 2; f'(1) = 1\) , then \(\lim_{x \to 1}\left[\frac{xf(1) - f(x)}{x - 1}\right] =\)
Answer: E
The limit is 1.
Add and subtract \(f(1)\) in the numerator:
\[ \begin{aligned} \frac{xf(1) - f(x)}{x - 1} &= \frac{(x - 1)f(1) - (f(x) - f(1))}{x - 1} \\ &= f(1) - \frac{f(x) - f(1)}{x - 1} \\ &\to f(1) - f'(1) = 2 - 1 = 1 \end{aligned} \]
The same result follows from L'Hospital's rule: the derivative of the numerator is \(f(1) - f'(x)\), which is 1 at \(x = 1\). A sign slip gives \(-1\).
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