Question

If $f(x) = \sqrt {25 - {x^2}} $ , then what is $\mathop {\lim }\limits_{x \to 1} \dfrac{{f(x) - f(1)}}{{x - 1}}$ equal to?
A. $\dfrac{1}{5}$
B. $\dfrac{1}{{24}}$
C. $\sqrt {24} $
D. $ - \dfrac{1}{{\sqrt {24} }}$

Answer 1

Hint: To solve this question we need to understand the concept of limit and differentiability. Here we use the differentiability at a point.
The function \[f(x)\] is said to be differentiable at the point \[x\; = \;a\;\] if the derivative \[f'(a)\;\] exists at every point in its domain. It is given by
\[f'(a)\; = \mathop {\lim }\limits_{x \to a} \dfrac{{f(x) - f(a)}}{{x - a}} \ldots (1)\]
If $f(x) = \sqrt {25 - {x^2}} $and \[\;a = 1\;\]then, from equation $(1)$ we get,
 $\mathop {\lim }\limits_{x \to 1} \dfrac{{f(x) - f(1)}}{{x - 1}} = f'(1)$
We have to find the derivative of $f(x) = \sqrt {25 - {x^2}} $ at point \[\;a = 1\;\].
Use the formulas of derivative;
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$\dfrac{d}{{dx}}(\sqrt x ) = \dfrac{1}{{2\sqrt x }}$
The value of $f'(1)$ is the required solution.

Complete step-by-step answer:
Given the function, $f(x) = \sqrt {25 - {x^2}} $ and use the differentiability at a point.
The function \[f(x)\] is said to be differentiable at the point \[x\; = \;a\;\] if the derivative \[f'(a)\;\] exists at every point in its domain. It is given by
\[f'(a)\; = \mathop {\lim }\limits_{x \to a} \dfrac{{f(x) - f(a)}}{{x - a}} \ldots (1)\]
Substitute \[\;a = 1\;\]into the equation $(1)$.
\[f'(1)\; = \mathop {\lim }\limits_{x \to a} \dfrac{{f(x) - f(1)}}{{x - 1}}\]
$\therefore $ Evaluate first order derivative of $f(x) = \sqrt {25 - {x^2}} $at point$1$.
$\dfrac{{d(\sqrt {25 - {x^2}} )}}{{dx}} = \dfrac{1}{{2\sqrt {25 - {x^2}} }}\dfrac{d}{{dx}}(25 - {x^2})$
$ \Rightarrow \dfrac{{d(\sqrt {25 - {x^2}} )}}{{dx}} = \dfrac{1}{{2\sqrt {25 - {x^2}} }}( - 2x)$
$ \Rightarrow \dfrac{{d(\sqrt {25 - {x^2}} )}}{{dx}} = - \dfrac{x}{{\sqrt {25 - {x^2}} }}$
$ \Rightarrow f'(x) = - \dfrac{x}{{\sqrt {25 - {x^2}} }}$
The value of $f'(1)$ is the required solution so, substitute $x = 1$ into $f'(x)$.
$ \Rightarrow f'(1) = - \dfrac{1}{{\sqrt {25 - {1^2}} }}$
$ \Rightarrow f'(1) = - \dfrac{1}{{\sqrt {24} }}$

Correct Answer: D. $ - \dfrac{1}{{\sqrt {24} }}$

Note:
Most important step is to compare $\mathop {\lim }\limits_{x \to 1} \dfrac{{f(x) - f(1)}}{{x - 1}}$ with the formula of differentiability \[f'(a)\; = \mathop {\lim }\limits_{x \to a} \dfrac{{f(x) - f(a)}}{{x - a}}\] and find $f'(1)$.
Here are the formulas of derivative given below,
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$\dfrac{d}{{dx}}(\sqrt x ) = \dfrac{1}{{2\sqrt x }}$
$\dfrac{d}{{dx}}{e^x} = {e^x}$
$\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = \log x$