To find the limit [tex]\(\lim_{x \to 2} \frac{x - \sqrt{8 - x^2}}{\sqrt{x^2 + 12} - 4}\)[/tex], let's evaluate the expression step-by-step.
1. Substitute [tex]\(x = 2\)[/tex] into the expression:
[tex]\[
\frac{2 - \sqrt{8 - 2^2}}{\sqrt{2^2 + 12} - 4}
\][/tex]
2. Simplify the expressions inside the square roots:
[tex]\[
2 - \sqrt{8 - 4} = 2 - \sqrt{4} = 2 - 2 = 0
\][/tex]
[tex]\[
\sqrt{2^2 + 12} - 4 = \sqrt{4 + 12} - 4 = \sqrt{16} - 4 = 4 - 4 = 0
\][/tex]
3. Since substituting [tex]\(x = 2\)[/tex] resulted in a [tex]\(\frac{0}{0}\)[/tex] indeterminate form, apply L'Hôpital's rule, which tells us to differentiate the numerator and the denominator and then take the limit again.
4. Differentiate the numerator and the denominator:
- Numerator: [tex]\(x - \sqrt{8 - x^2}\)[/tex]
- The derivative of [tex]\(x\)[/tex] is 1.
- The derivative of [tex]\(-\sqrt{8 - x^2}\)[/tex] can be found using the chain rule:
[tex]\[
\frac{d}{dx}\left(-\sqrt{8 - x^2}\right) = -\frac{1}{2}(8 - x^2)^{-\frac{1}{2}} \cdot (-2x) = \frac{x}{\sqrt{8 - x^2}}
\][/tex]
Therefore, the derivative of the numerator:
[tex]\[
1 + \frac{x}{\sqrt{8 - x^2}}
\][/tex]
- Denominator: [tex]\(\sqrt{x^2 + 12} - 4\)[/tex]
- The derivative of [tex]\(\sqrt{x^2 + 12}\)[/tex] is:
[tex]\[
\frac{d}{dx}\left(\sqrt{x^2 + 12}\right) = \frac{1}{2}(x^2 + 12)^{-\frac{1}{2}} \cdot 2x = \frac{x}{\sqrt{x^2 + 12}}
\][/tex]
Therefore, the derivative of the denominator:
[tex]\[
\frac{x}{\sqrt{x^2 + 12}}
\][/tex]
5. Construct the new limit using the derivatives:
[tex]\[
\lim_{x \to 2} \frac{1 + \frac{x}{\sqrt{8 - x^2}}}{\frac{x}{\sqrt{x^2 + 12}}}
\][/tex]
6. Substitute [tex]\(x = 2\)[/tex] into the new expression:
[tex]\[
\frac{1 + \frac{2}{\sqrt{8 - 4}}}{\frac{2}{\sqrt{4 + 12}}} = \frac{1 + \frac{2}{2}}{\frac{2}{4}} = \frac{1 + 1}{\frac{1}{2}} = \frac{2}{\frac{1}{2}} = 2 \times 2 = 4
\][/tex]
Thus, the limit is:
[tex]\[
\lim_{x \to 2} \frac{x - \sqrt{8 - x^2}}{\sqrt{x^2 + 12} - 4} = 4
\][/tex]
