IDNLearn.com helps you find the answers you need quickly and efficiently. Our experts provide timely and precise responses to help you understand and solve any issue you face.
Sagot :
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]
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]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.