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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]
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