Answer :
To evaluate the limit
[tex]\[ \lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}}, \][/tex]
we need to handle both the numerator and the denominator carefully as [tex]\( x \)[/tex] approaches 1.
### Step-by-Step Solution:
Step 1: Substitute [tex]\( x = 1 \)[/tex] directly.
Substituting [tex]\( x = 1 \)[/tex] into the expression, we get:
[tex]\[ x - \sqrt{2 - x^2} = 1 - \sqrt{2 - 1^2} = 1 - \sqrt{1} = 1 - 1 = 0, \][/tex]
[tex]\[ 2x - \sqrt{2 + 2x^2} = 2(1) - \sqrt{2 + 2(1^2)} = 2 - \sqrt{2 + 2} = 2 - \sqrt{4} = 2 - 2 = 0. \][/tex]
Both the numerator [tex]\( x - \sqrt{2 - x^2} \)[/tex] and the denominator [tex]\( 2x - \sqrt{2 + 2x^2} \)[/tex] approach 0 as [tex]\( x \)[/tex] approaches 1, resulting in a [tex]\( \frac{0}{0} \)[/tex] indeterminate form. Thus, we can apply L'Hôpital's Rule.
Step 2: Apply L'Hôpital's Rule.
L'Hôpital's Rule states that if we have a limit of the form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], we can take the derivatives of the numerator and the denominator separately and then re-evaluate the limit. So, we need to find the derivatives of the numerator and the denominator.
Step 3: Differentiate the numerator and the denominator.
Let's differentiate the numerator [tex]\( x - \sqrt{2 - x^2} \)[/tex]:
[tex]\[ \frac{d}{dx} \left( x - \sqrt{2 - x^2} \right) = 1 - \frac{d}{dx} \left( \sqrt{2 - x^2} \right). \][/tex]
Using the chain rule for the square root term:
[tex]\[ \frac{d}{dx} \left( \sqrt{2 - x^2} \right) = \frac{1}{2\sqrt{2 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{2 - x^2}}. \][/tex]
So, the derivative of the numerator is:
[tex]\[ 1 - \frac{-x}{\sqrt{2 - x^2}} = 1 + \frac{x}{\sqrt{2 - x^2}}. \][/tex]
Next, let's differentiate the denominator [tex]\( 2x - \sqrt{2 + 2x^2} \)[/tex]:
[tex]\[ \frac{d}{dx} \left( 2x - \sqrt{2 + 2x^2} \right) = 2 - \frac{d}{dx} \left( \sqrt{2 + 2x^2} \right). \][/tex]
Again, using the chain rule for the square root term:
[tex]\[ \frac{d}{dx} \left( \sqrt{2 + 2x^2} \right) = \frac{1}{2\sqrt{2 + 2x^2}} \cdot (4x) = \frac{2x}{\sqrt{2 + 2x^2}}. \][/tex]
So, the derivative of the denominator is:
[tex]\[ 2 - \frac{2x}{\sqrt{2 + 2x^2}}. \][/tex]
Step 4: Re-evaluate the limit using these derivatives.
Thus, our new limit is:
[tex]\[ \lim_{x \to 1} \frac{1 + \frac{x}{\sqrt{2 - x^2}}}{2 - \frac{2x}{\sqrt{2 + 2x^2}}}. \][/tex]
Now, substituting [tex]\( x = 1 \)[/tex] into this new expression:
For the numerator:
[tex]\[ 1 + \frac{1}{\sqrt{2 - 1}} = 1 + \frac{1}{\sqrt{1}} = 1 + 1 = 2. \][/tex]
For the denominator:
[tex]\[ 2 - \frac{2(1)}{\sqrt{2 + 2(1)^2}} = 2 - \frac{2}{\sqrt{4}} = 2 - \frac{2}{2} = 2 - 1 = 1. \][/tex]
So, the limit is:
[tex]\[ \frac{2}{1} = 2. \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}} = 2. \][/tex]
[tex]\[ \lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}}, \][/tex]
we need to handle both the numerator and the denominator carefully as [tex]\( x \)[/tex] approaches 1.
### Step-by-Step Solution:
Step 1: Substitute [tex]\( x = 1 \)[/tex] directly.
Substituting [tex]\( x = 1 \)[/tex] into the expression, we get:
[tex]\[ x - \sqrt{2 - x^2} = 1 - \sqrt{2 - 1^2} = 1 - \sqrt{1} = 1 - 1 = 0, \][/tex]
[tex]\[ 2x - \sqrt{2 + 2x^2} = 2(1) - \sqrt{2 + 2(1^2)} = 2 - \sqrt{2 + 2} = 2 - \sqrt{4} = 2 - 2 = 0. \][/tex]
Both the numerator [tex]\( x - \sqrt{2 - x^2} \)[/tex] and the denominator [tex]\( 2x - \sqrt{2 + 2x^2} \)[/tex] approach 0 as [tex]\( x \)[/tex] approaches 1, resulting in a [tex]\( \frac{0}{0} \)[/tex] indeterminate form. Thus, we can apply L'Hôpital's Rule.
Step 2: Apply L'Hôpital's Rule.
L'Hôpital's Rule states that if we have a limit of the form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], we can take the derivatives of the numerator and the denominator separately and then re-evaluate the limit. So, we need to find the derivatives of the numerator and the denominator.
Step 3: Differentiate the numerator and the denominator.
Let's differentiate the numerator [tex]\( x - \sqrt{2 - x^2} \)[/tex]:
[tex]\[ \frac{d}{dx} \left( x - \sqrt{2 - x^2} \right) = 1 - \frac{d}{dx} \left( \sqrt{2 - x^2} \right). \][/tex]
Using the chain rule for the square root term:
[tex]\[ \frac{d}{dx} \left( \sqrt{2 - x^2} \right) = \frac{1}{2\sqrt{2 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{2 - x^2}}. \][/tex]
So, the derivative of the numerator is:
[tex]\[ 1 - \frac{-x}{\sqrt{2 - x^2}} = 1 + \frac{x}{\sqrt{2 - x^2}}. \][/tex]
Next, let's differentiate the denominator [tex]\( 2x - \sqrt{2 + 2x^2} \)[/tex]:
[tex]\[ \frac{d}{dx} \left( 2x - \sqrt{2 + 2x^2} \right) = 2 - \frac{d}{dx} \left( \sqrt{2 + 2x^2} \right). \][/tex]
Again, using the chain rule for the square root term:
[tex]\[ \frac{d}{dx} \left( \sqrt{2 + 2x^2} \right) = \frac{1}{2\sqrt{2 + 2x^2}} \cdot (4x) = \frac{2x}{\sqrt{2 + 2x^2}}. \][/tex]
So, the derivative of the denominator is:
[tex]\[ 2 - \frac{2x}{\sqrt{2 + 2x^2}}. \][/tex]
Step 4: Re-evaluate the limit using these derivatives.
Thus, our new limit is:
[tex]\[ \lim_{x \to 1} \frac{1 + \frac{x}{\sqrt{2 - x^2}}}{2 - \frac{2x}{\sqrt{2 + 2x^2}}}. \][/tex]
Now, substituting [tex]\( x = 1 \)[/tex] into this new expression:
For the numerator:
[tex]\[ 1 + \frac{1}{\sqrt{2 - 1}} = 1 + \frac{1}{\sqrt{1}} = 1 + 1 = 2. \][/tex]
For the denominator:
[tex]\[ 2 - \frac{2(1)}{\sqrt{2 + 2(1)^2}} = 2 - \frac{2}{\sqrt{4}} = 2 - \frac{2}{2} = 2 - 1 = 1. \][/tex]
So, the limit is:
[tex]\[ \frac{2}{1} = 2. \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}} = 2. \][/tex]
