Answer :
Certainly! Let's evaluate the limit:
[tex]\[ \lim_{x \rightarrow 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}} \][/tex]
### Step-by-Step Solution:
1. Substitute [tex]\( x = 1 \)[/tex]:
First, we will substitute [tex]\( x = 1 \)[/tex] directly into the expression to see if it results in an indeterminate form.
[tex]\[ \frac{1 - \sqrt{2 - 1^2}}{2 \cdot 1 - \sqrt{2 + 2 \cdot 1^2}} = \frac{1 - \sqrt{1}}{2 - \sqrt{4}} = \frac{1 - 1}{2 - 2} = \frac{0}{0} \][/tex]
Since the direct substitution produces [tex]\( \frac{0}{0} \)[/tex], this is an indeterminate form, and we need to use algebraic manipulation, such as rationalization or L'Hôpital's Rule, to evaluate the limit.
2. Rationalize the Numerator and Denominator:
To handle the square roots, let's try rationalizing the numerator and the denominator.
The numerator is:
[tex]\[ x - \sqrt{2 - x^2} \][/tex]
To rationalize the numerator, multiply by the conjugate:
[tex]\[ \frac{x - \sqrt{2 - x^2}}{\sqrt{2 - x^2} + x} \][/tex]
So the numerator becomes:
[tex]\[ \left( x - \sqrt{2 - x^2} \right) \cdot \left( \sqrt{2 - x^2} + x \right) = x \cdot \sqrt{2 - x^2} + x^2 - (2 - x^2) = x \sqrt{2 - x^2} + x^2 - 2 + x^2 = x \sqrt{2 - x^2} + 2x^2 - 2 \][/tex]
The denominator is:
[tex]\[ 2x - \sqrt{2 + 2x^2} \][/tex]
To rationalize the denominator, multiply by the conjugate:
[tex]\[ \frac{2x - \sqrt{2 + 2x^2}}{2x + \sqrt{2 + 2x^2}} \][/tex]
So the denominator becomes:
[tex]\[ \left( 2x - \sqrt{2 + 2x^2} \right) \cdot \left( 2x + \sqrt{2 + 2x^2} \right) = 4x^2 - (2 + 2x^2) = 4x^2 - 2 - 2x^2 = 2x^2 - 2 \][/tex]
3. Combine the Results:
Now, instead of directly doing the rationalization (for ease):
After simplification and considering rationalization indeed sets up to easier limits calculable via L'Hopital Rule or evaluations:
4. Applying L'Hopital's Rule:
Since [tex]\( \frac{0}{0} \)[/tex] is an indeterminate form, apply L'Hopital's Rule by taking derivatives of the numerator and the denominator:
Let [tex]\( f(x) = x - \sqrt{2 - x^2} \)[/tex] and [tex]\( g(x) = 2x - \sqrt{2 + 2x^2} \)[/tex].
Compute [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = 1 - \frac{d}{dx} (\sqrt{2 - x^2}) = 1 - \frac{1}{2\sqrt{2 - x^2}} \cdot (-2x) = 1 + \frac{x}{\sqrt{2 - x^2}} \][/tex]
Compute [tex]\( g'(x) \)[/tex]:
[tex]\[ g'(x) = 2 - \frac{d}{dx} (\sqrt{2 + 2x^2}) = 2 - \frac{1}{2\sqrt{2 + 2x^2}} \cdot (4x) = 2 - \frac{2x}{\sqrt{2 + 2x^2}} \][/tex]
Apply L'Hopital's Rule:
[tex]\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{1 + \frac{x}{\sqrt{2 - x^2}}}{2 - \frac{2x}{\sqrt{2 + 2x^2}}} \][/tex]
Substitute [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{1 + \frac{1}{\sqrt{1}}}{2 - \frac{2 \cdot 1}{\sqrt{4}}} = \frac{1 + 1}{2 - 1} = \frac{2}{1} = 2 \][/tex]
### The final answer is:
[tex]\[ \boxed{2} \][/tex]
[tex]\[ \lim_{x \rightarrow 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}} \][/tex]
### Step-by-Step Solution:
1. Substitute [tex]\( x = 1 \)[/tex]:
First, we will substitute [tex]\( x = 1 \)[/tex] directly into the expression to see if it results in an indeterminate form.
[tex]\[ \frac{1 - \sqrt{2 - 1^2}}{2 \cdot 1 - \sqrt{2 + 2 \cdot 1^2}} = \frac{1 - \sqrt{1}}{2 - \sqrt{4}} = \frac{1 - 1}{2 - 2} = \frac{0}{0} \][/tex]
Since the direct substitution produces [tex]\( \frac{0}{0} \)[/tex], this is an indeterminate form, and we need to use algebraic manipulation, such as rationalization or L'Hôpital's Rule, to evaluate the limit.
2. Rationalize the Numerator and Denominator:
To handle the square roots, let's try rationalizing the numerator and the denominator.
The numerator is:
[tex]\[ x - \sqrt{2 - x^2} \][/tex]
To rationalize the numerator, multiply by the conjugate:
[tex]\[ \frac{x - \sqrt{2 - x^2}}{\sqrt{2 - x^2} + x} \][/tex]
So the numerator becomes:
[tex]\[ \left( x - \sqrt{2 - x^2} \right) \cdot \left( \sqrt{2 - x^2} + x \right) = x \cdot \sqrt{2 - x^2} + x^2 - (2 - x^2) = x \sqrt{2 - x^2} + x^2 - 2 + x^2 = x \sqrt{2 - x^2} + 2x^2 - 2 \][/tex]
The denominator is:
[tex]\[ 2x - \sqrt{2 + 2x^2} \][/tex]
To rationalize the denominator, multiply by the conjugate:
[tex]\[ \frac{2x - \sqrt{2 + 2x^2}}{2x + \sqrt{2 + 2x^2}} \][/tex]
So the denominator becomes:
[tex]\[ \left( 2x - \sqrt{2 + 2x^2} \right) \cdot \left( 2x + \sqrt{2 + 2x^2} \right) = 4x^2 - (2 + 2x^2) = 4x^2 - 2 - 2x^2 = 2x^2 - 2 \][/tex]
3. Combine the Results:
Now, instead of directly doing the rationalization (for ease):
After simplification and considering rationalization indeed sets up to easier limits calculable via L'Hopital Rule or evaluations:
4. Applying L'Hopital's Rule:
Since [tex]\( \frac{0}{0} \)[/tex] is an indeterminate form, apply L'Hopital's Rule by taking derivatives of the numerator and the denominator:
Let [tex]\( f(x) = x - \sqrt{2 - x^2} \)[/tex] and [tex]\( g(x) = 2x - \sqrt{2 + 2x^2} \)[/tex].
Compute [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = 1 - \frac{d}{dx} (\sqrt{2 - x^2}) = 1 - \frac{1}{2\sqrt{2 - x^2}} \cdot (-2x) = 1 + \frac{x}{\sqrt{2 - x^2}} \][/tex]
Compute [tex]\( g'(x) \)[/tex]:
[tex]\[ g'(x) = 2 - \frac{d}{dx} (\sqrt{2 + 2x^2}) = 2 - \frac{1}{2\sqrt{2 + 2x^2}} \cdot (4x) = 2 - \frac{2x}{\sqrt{2 + 2x^2}} \][/tex]
Apply L'Hopital's Rule:
[tex]\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{1 + \frac{x}{\sqrt{2 - x^2}}}{2 - \frac{2x}{\sqrt{2 + 2x^2}}} \][/tex]
Substitute [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{1 + \frac{1}{\sqrt{1}}}{2 - \frac{2 \cdot 1}{\sqrt{4}}} = \frac{1 + 1}{2 - 1} = \frac{2}{1} = 2 \][/tex]
### The final answer is:
[tex]\[ \boxed{2} \][/tex]
