Evaluating the Limit Using Derivative Definition

The problem asks us to evaluate a specific limit expression involving the function \({\rm{F}}({\rm{x}}) = \sqrt{9 - {{\rm{x}}^2}}\). The limit expression is given by: \[ \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{F}}\left( {\rm{x}} \right) - {\rm{F}}\left( 1 \right)}}{{{\rm{x}} - 1}} \]

This form of the limit is precisely the definition of the derivative of the function \({\rm{F}}({\rm{x}})\) at the point \({\rm{x}} = 1\). Recall the definition of the derivative of a function \(f({\rm{x}})\) at a point \(a\): \[ f'(a) = \mathop {\lim }\limits_{{\rm{x}} \to a} \frac{{f\left( {\rm{x}} \right) - f\left( a \right)}}{{{\rm{x}} - a}} \] In this case, \(f({\rm{x}}) = {\rm{F}}({\rm{x}}) = \sqrt{9 - {{\rm{x}}^2}}\) and \(a = 1\).

Therefore, the value of the given limit is equal to the derivative of \({\rm{F}}({\rm{x}})\) evaluated at \({\rm{x}} = 1\), i.e., \({\rm{F}}'(1)\).

Calculating the Derivative of \({\rm{F}}({\rm{x}})\)

We need to find the derivative of \({\rm{F}}({\rm{x}}) = \sqrt{9 - {{\rm{x}}^2}}\) with respect to \({\rm{x}}\). We can use the chain rule for differentiation.

Let \(u = 9 - {{\rm{x}}^2}\). Then \({\rm{F}}({\rm{x}}) = \sqrt{u} = u^{1/2}\).

Now, we find the derivatives of \(u\) with respect to \({\rm{x}}\) and \({\rm{F}}\) with respect to \(u\):

• Derivative of \(u\) with respect to \({\rm{x}}\): \[ \frac{du}{d{\rm{x}}} = \frac{d}{d{\rm{x}}}(9 - {{\rm{x}}^2}) = 0 - 2{\rm{x}} = -2{\rm{x}} \]
• Derivative of \({\rm{F}}\) with respect to \(u\): \[ \frac{d{\rm{F}}}{du} = \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{1/2 - 1} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} \]

Using the chain rule, \({\rm{F}}'({\rm{x}}) = \frac{d{\rm{F}}}{du} \cdot \frac{du}{d{\rm{x}}}\): \[ {\rm{F}}'({\rm{x}}) = \left( \frac{1}{2\sqrt{u}} \right) \cdot (-2{\rm{x}}) \] Substitute \(u = 9 - {{\rm{x}}^2}\) back into the expression for \({\rm{F}}'({\rm{x}})\): \[ {\rm{F}}'({\rm{x}}) = \frac{1}{2\sqrt{9 - {{\rm{x}}^2}}} \cdot (-2{\rm{x}}) = \frac{-2{\rm{x}}}{2\sqrt{9 - {{\rm{x}}^2}}} = \frac{-{\rm{x}}}{\sqrt{9 - {{\rm{x}}^2}}} \]

Evaluating \({\rm{F}}'(1)\)

Now we need to evaluate the derivative \({\rm{F}}'({\rm{x}})\) at \({\rm{x}} = 1\). Substitute \({\rm{x}} = 1\) into the expression for \({\rm{F}}'({\rm{x}})\): \[ {\rm{F}}'(1) = \frac{-1}{\sqrt{9 - 1^2}} = \frac{-1}{\sqrt{9 - 1}} = \frac{-1}{\sqrt{8}} \]

Simplifying the Result

The value \( \frac{-1}{\sqrt{8}} \) can be simplified. We know that \( \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2} \).

So, \( {\rm{F}}'(1) = \frac{-1}{2\sqrt{2}} \).

Conclusion

The value of the limit \( \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{F}}\left( {\rm{x}} \right) - {\rm{F}}\left( 1 \right)}}{{{\rm{x}} - 1}} \) is equal to \({\rm{F}}'(1)\), which we calculated to be \( \frac{-1}{2\sqrt{2}} \).

Revision Table: Key Concepts

This problem utilized the definition of the derivative and the chain rule for differentiation.

• Limit Definition of Derivative: \( f'(a) = \mathop {\lim }\limits_{{\rm{x}} \to a} \frac{{f\left( {\rm{x}} \right) - f\left( a \right)}}{{{\rm{x}} - a}} \)
• Chain Rule: If \(y = f(u)\) and \(u = g({\rm{x}})\), then \( \frac{dy}{d{\rm{x}}} = \frac{dy}{du} \cdot \frac{du}{d{\rm{x}}} \)
• Derivative of Power Function: \( \frac{d}{d{\rm{x}}}({{\rm{x}}^n}) = n{{\rm{x}}^{n-1}} \)
• Derivative of Constant: \( \frac{d}{d{\rm{x}}}(c) = 0 \)

Additional Information: Domain of F(x)

The function \({\rm{F}}({\rm{x}}) = \sqrt{9 - {{\rm{x}}^2}}\) is defined when \(9 - {{\rm{x}}^2} \ge 0\), which means \( {{\rm{x}}^2} \le 9 \). This inequality holds for \(-3 \le {\rm{x}} \le 3\). The point \({\rm{x}} = 1\) is within this domain, and the derivative is well-defined at \({\rm{x}}=1\) because \(9 - 1^2 = 8 > 0\).