Question

If $f(x)=\sqrt{10-x}$, then $\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}$ is equal to:

• A. \(3\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{6}\)
• D. \(-\frac{1}{6}\)
• E. \(\frac{3}{2}\)

Hint:

Limits of the form \(\frac{f(x)-f(a)}{x-a}\) directly represent \(f'(a)\). Use differentiation instead of expanding the limit manually.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given limit expression, \( \lim_{x\to a} \frac{f(x)-f(a)}{x-a} \), is the definition of the derivative of the function \( f(x) \) at the point \( x = a \). In this problem, \( a=1 \). Therefore, we need to find the derivative of \( f(x) \) and evaluate it at \( x=1 \), i.e., find \( f'(1) \).
Step 2: Key Formula or Approach:
The derivative of a function \( f(x) = x^n \) is given by \( f'(x) = nx^{n-1} \). We will also use the chain rule, which states that if \( h(x) = g(f(x)) \), then \( h'(x) = g'(f(x)) \cdot f'(x) \).
The function is \( f(x) = \sqrt{10-x} = (10-x)^{1/2} \).
Step 3: Detailed Explanation:
First, we find the derivative of \( f(x) \) with respect to \( x \).
\[ f(x) = (10-x)^{1/2} \] Using the power rule combined with the chain rule:
\[ f'(x) = \frac{1}{2}(10-x)^{\frac{1}{2}-1} \cdot \frac{d}{dx}(10-x) \] \[ f'(x) = \frac{1}{2}(10-x)^{-1/2} \cdot (-1) \] \[ f'(x) = -\frac{1}{2\sqrt{10-x}} \] Now, we evaluate this derivative at \( x=1 \).
\[ f'(1) = -\frac{1}{2\sqrt{10-1}} \] \[ f'(1) = -\frac{1}{2\sqrt{9}} \] \[ f'(1) = -\frac{1}{2 \cdot 3} \] \[ f'(1) = -\frac{1}{6} \] Step 4: Final Answer:
The value of the limit is equal to the value of the derivative at \( x=1 \), which is \( -\frac{1}{6} \). This corresponds to option (D).