Question

Let $f(x)=-\sqrt{49-x^{2}}$. Then $\lim_{x\rightarrow1}\frac{f(x)-f(1)}{x-1}$ is equal to

• A. $\frac{1}{3\sqrt{3}}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{4\sqrt{3}}$
• E. $\frac{1}{5\sqrt{3}}$

Hint:

Math Tip: Whenever you see a limit in the exact form of $\frac{f(x)-f(a)}{x-a}$, never try to solve it using algebraic limits or L'Hôpital's rule directly. Recognize it as the definition of the derivative $f'(a)$ and just differentiate the function!

Answer 1

The Correct Option is D

Concept:
Calculus - First Principle of Derivatives.
The limit definition of a derivative at a specific point $x = a$ is: $$ f'(a) = \lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} $$
Step 1: Identify the objective.
By comparing the given limit to the formal definition, we see that it exactly represents the derivative of $f(x)$ evaluated at $x = 1$. $$ \text{Limit} = f'(1) $$
Step 2: Find the derivative of the function.
Given function: $f(x) = -(49-x^2)^{\frac{1}{2}}$. Apply the chain rule to differentiate with respect to $x$: $$ f'(x) = -\frac{1}{2}(49-x^2)^{-\frac{1}{2}} \cdot \frac{d}{dx}(49-x^2) $$ $$ f'(x) = -\frac{1}{2\sqrt{49-x^2}} \cdot (-2x) $$
Step 3: Simplify the derivative expression.
Cancel out the $-2$ terms in the numerator and denominator: $$ f'(x) = \frac{x}{\sqrt{49-x^2}} $$
Step 4: Evaluate the derivative at $x = 1$.
Substitute $x = 1$ into the simplified derivative: $$ f'(1) = \frac{1}{\sqrt{49 - (1)^2}} $$ $$ f'(1) = \frac{1}{\sqrt{48}} $$
Step 5: Simplify the surd to match the options.
Factor 48 into a perfect square: $$ f'(1) = \frac{1}{\sqrt{16 \times 3}} $$ $$ f'(1) = \frac{1}{4\sqrt{3}} $$