Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To evaluate the limit [tex]\(\lim _{x \rightarrow 1} \frac{x-\sqrt{2-x^2}}{2x-\sqrt{2+2x^2}}\)[/tex], we need to carefully examine the function and determine its behavior as [tex]\( x \)[/tex] approaches 1. Here is a step-by-step approach to solve this problem:
1. Substitute [tex]\( x = 1 \)[/tex] into the function:
[tex]\[ \frac{1-\sqrt{2-1^2}}{2\cdot1-\sqrt{2+2\cdot1^2}} = \frac{1-\sqrt{2-1}}{2-\sqrt{2+2}} = \frac{1-\sqrt{1}}{2-\sqrt{4}} \][/tex]
Simplifying the square roots, we get:
[tex]\[ \frac{1-1}{2-2} = \frac{0}{0} \][/tex]
This gives us the indeterminate form [tex]\( \frac{0}{0} \)[/tex].
2. Apply L'Hôpital's Rule:
Since we have an indeterminate form of [tex]\( \frac{0}{0} \)[/tex], we can use L'Hôpital's Rule, which states that for limits of the form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], the limit of the ratio of functions is the limit of the ratio of their derivatives. L'Hôpital's Rule means we need to differentiate the numerator and the denominator separately and then take the limit again.
Let's differentiate the numerator [tex]\( x - \sqrt{2 - x^2} \)[/tex] and the denominator [tex]\( 2x - \sqrt{2 + 2x^2} \)[/tex]:
- Numerator: Differentiate [tex]\( x - \sqrt{2 - x^2} \)[/tex]
[tex]\[ \text{Let } y = x - \sqrt{2 - x^2} \Rightarrow \frac{dy}{dx} = 1 - \frac{1}{2\sqrt{2 - x^2}} \cdot (-2x) \][/tex]
Simplify the derivative in the numerator:
[tex]\[ \frac{dy}{dx} = 1 + \frac{x}{\sqrt{2 - x^2}} \][/tex]
- Denominator: Differentiate [tex]\( 2x - \sqrt{2 + 2x^2} \)[/tex]
[tex]\[ \text{Let } z = 2x - \sqrt{2 + 2x^2} \Rightarrow \frac{dz}{dx} = 2 - \frac{1}{2\sqrt{2 + 2x^2}} \cdot (4x) \][/tex]
Simplify the derivative in the denominator:
[tex]\[ \frac{dz}{dx} = 2 - \frac{2x}{\sqrt{2 + 2x^2}} \][/tex]
3. Substitute [tex]\( x = 1 \)[/tex] into the derivatives:
[tex]\[ \lim_{x \to 1} \frac{\frac{dy}{dx}}{\frac{dz}{dx}} = \lim_{x \to 1} \frac{1 + \frac{x}{\sqrt{2 - x^2}}}{2 - \frac{2x}{\sqrt{2 + 2x^2}}} \][/tex]
Substitute [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{1 + \frac{1}{\sqrt{2 - 1}}}{2 - \frac{2 \cdot 1}{\sqrt{2 + 2 \cdot 1^2}}} = \frac{1 + \frac{1}{\sqrt{1}}}{2 - \frac{2}{\sqrt{4}}} = \frac{1 + 1}{2 - 1} = \frac{2}{1} = 2 \][/tex]
The limit evaluates to 2. Hence, the solution is:
[tex]\[ \lim _{x \rightarrow 1} \frac{x-\sqrt{2-x^2}}{2 x-\sqrt{2+2 x^2}} = 2 \][/tex]
1. Substitute [tex]\( x = 1 \)[/tex] into the function:
[tex]\[ \frac{1-\sqrt{2-1^2}}{2\cdot1-\sqrt{2+2\cdot1^2}} = \frac{1-\sqrt{2-1}}{2-\sqrt{2+2}} = \frac{1-\sqrt{1}}{2-\sqrt{4}} \][/tex]
Simplifying the square roots, we get:
[tex]\[ \frac{1-1}{2-2} = \frac{0}{0} \][/tex]
This gives us the indeterminate form [tex]\( \frac{0}{0} \)[/tex].
2. Apply L'Hôpital's Rule:
Since we have an indeterminate form of [tex]\( \frac{0}{0} \)[/tex], we can use L'Hôpital's Rule, which states that for limits of the form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], the limit of the ratio of functions is the limit of the ratio of their derivatives. L'Hôpital's Rule means we need to differentiate the numerator and the denominator separately and then take the limit again.
Let's differentiate the numerator [tex]\( x - \sqrt{2 - x^2} \)[/tex] and the denominator [tex]\( 2x - \sqrt{2 + 2x^2} \)[/tex]:
- Numerator: Differentiate [tex]\( x - \sqrt{2 - x^2} \)[/tex]
[tex]\[ \text{Let } y = x - \sqrt{2 - x^2} \Rightarrow \frac{dy}{dx} = 1 - \frac{1}{2\sqrt{2 - x^2}} \cdot (-2x) \][/tex]
Simplify the derivative in the numerator:
[tex]\[ \frac{dy}{dx} = 1 + \frac{x}{\sqrt{2 - x^2}} \][/tex]
- Denominator: Differentiate [tex]\( 2x - \sqrt{2 + 2x^2} \)[/tex]
[tex]\[ \text{Let } z = 2x - \sqrt{2 + 2x^2} \Rightarrow \frac{dz}{dx} = 2 - \frac{1}{2\sqrt{2 + 2x^2}} \cdot (4x) \][/tex]
Simplify the derivative in the denominator:
[tex]\[ \frac{dz}{dx} = 2 - \frac{2x}{\sqrt{2 + 2x^2}} \][/tex]
3. Substitute [tex]\( x = 1 \)[/tex] into the derivatives:
[tex]\[ \lim_{x \to 1} \frac{\frac{dy}{dx}}{\frac{dz}{dx}} = \lim_{x \to 1} \frac{1 + \frac{x}{\sqrt{2 - x^2}}}{2 - \frac{2x}{\sqrt{2 + 2x^2}}} \][/tex]
Substitute [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{1 + \frac{1}{\sqrt{2 - 1}}}{2 - \frac{2 \cdot 1}{\sqrt{2 + 2 \cdot 1^2}}} = \frac{1 + \frac{1}{\sqrt{1}}}{2 - \frac{2}{\sqrt{4}}} = \frac{1 + 1}{2 - 1} = \frac{2}{1} = 2 \][/tex]
The limit evaluates to 2. Hence, the solution is:
[tex]\[ \lim _{x \rightarrow 1} \frac{x-\sqrt{2-x^2}}{2 x-\sqrt{2+2 x^2}} = 2 \][/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
