Discussion
If \(f(x) = \sqrt{10-x}\), then \(\lim_{x \to 1} \frac{f(x)-f(1)}{x-1}\) is equal to
Answer: D
The limit is \(-\frac{1}{6}\).
The limit is the definition of the derivative \(f'(1)\):
\[ \begin{aligned} f'(x) &= \frac{-1}{2\sqrt{10 - x}} \\ f'(1) &= \frac{-1}{2\sqrt{9}} = -\frac{1}{6} \end{aligned} \]
The minus sign comes from the chain rule, since the derivative of \(10 - x\) is -1. \(\frac{1}{6}\) drops that sign, and 3 is \(f(1)\), the value of the function rather than its rate of change.
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