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Sagot :
To find the limit [tex]\(\lim _{x \rightarrow 2} \frac{\sqrt{2 + \sqrt{2 + x}} - 2}{x - 2}\)[/tex], let's go through it step by step.
1. Direct Substitution:
First, if we directly substitute [tex]\( x = 2 \)[/tex] into the function [tex]\(\frac{\sqrt{2 + \sqrt{2 + x}} - 2}{x - 2}\)[/tex]:
[tex]\[ \frac{\sqrt{2 + \sqrt{2 + 2}} - 2}{2 - 2} = \frac{\sqrt{2 + \sqrt{4}} - 2}{0} = \frac{\sqrt{2 + 2} - 2}{0} = \frac{\sqrt{4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0} \][/tex]
This produces an indeterminate form [tex]\(\frac{0}{0}\)[/tex], which means further analysis is needed.
2. L'Hopital's Rule:
Since the direct substitution led to an indeterminate form [tex]\(\frac{0}{0}\)[/tex], we can apply L'Hopital's Rule, which states that if [tex]\(\lim_{x \to a} \frac{f(x)}{g(x)}\)[/tex] yields an indeterminate form, the limit can be evaluated as:
[tex]\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \][/tex]
provided the limit on the right exists. Here:
[tex]\[ f(x) = \sqrt{2 + \sqrt{2 + x}} - 2, \quad g(x) = x - 2 \][/tex]
3. Differentiate the Numerator and Denominator:
Differentiate [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ f(x) = \sqrt{2 + \sqrt{2 + x}} - 2 \][/tex]
The derivative [tex]\( f'(x) \)[/tex] involves the chain rule. Let [tex]\( u = 2 + \sqrt{2 + x} \)[/tex]. Then,
[tex]\[ f(x) = \sqrt{u} - 2 \][/tex]
So,
[tex]\[ f'(x) = \frac{d}{dx} (\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \][/tex]
Now,
[tex]\[ \frac{du}{dx} = \frac{d}{dx} (2 + \sqrt{2 + x}) = \frac{d}{dx} (\sqrt{2 + x}) = \frac{1}{2\sqrt{2 + x}} \][/tex]
Putting it all together,
[tex]\[ f'(x) = \frac{1}{2\sqrt{2 + \sqrt{2 + x}}} \cdot \frac{1}{2\sqrt{2 + x}} \][/tex]
Simplify:
[tex]\[ f'(x) = \frac{1}{4\sqrt{2 + \sqrt{2 + x}} \sqrt{2 + x}} \][/tex]
The derivative of the denominator [tex]\( g(x) = x - 2 \)[/tex] is simply:
[tex]\[ g'(x) = 1 \][/tex]
4. Apply L'Hopital's Rule:
Now, we can apply L'Hopital's Rule:
[tex]\[ \lim_{x \to 2} \frac{\sqrt{2 + \sqrt{2 + x}} - 2}{x - 2} = \lim_{x \to 2} \frac{\frac{1}{4\sqrt{2 + \sqrt{2 + x}} \sqrt{2 + x}}}{1} \][/tex]
5. Evaluate the Limit:
Substitute [tex]\( x = 2 \)[/tex] in the simplified derivative:
[tex]\[ \lim_{x \to 2} \frac{1}{4\sqrt{2 + \sqrt{2 + x}} \sqrt{2 + x}} = \frac{1}{4\sqrt{2 + \sqrt{4}} \sqrt{4}} = \frac{1}{4\sqrt{2 + 2} \cdot 2} = \frac{1}{4 \cdot 2 \cdot 2} = \frac{1}{16} \][/tex]
6. Final Answer:
Therefore, the limit is:
[tex]\[ \lim _{x \rightarrow 2} \frac{\sqrt{2 + \sqrt{2 + x}} - 2}{x - 2} = \frac{1}{16} \][/tex]
1. Direct Substitution:
First, if we directly substitute [tex]\( x = 2 \)[/tex] into the function [tex]\(\frac{\sqrt{2 + \sqrt{2 + x}} - 2}{x - 2}\)[/tex]:
[tex]\[ \frac{\sqrt{2 + \sqrt{2 + 2}} - 2}{2 - 2} = \frac{\sqrt{2 + \sqrt{4}} - 2}{0} = \frac{\sqrt{2 + 2} - 2}{0} = \frac{\sqrt{4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0} \][/tex]
This produces an indeterminate form [tex]\(\frac{0}{0}\)[/tex], which means further analysis is needed.
2. L'Hopital's Rule:
Since the direct substitution led to an indeterminate form [tex]\(\frac{0}{0}\)[/tex], we can apply L'Hopital's Rule, which states that if [tex]\(\lim_{x \to a} \frac{f(x)}{g(x)}\)[/tex] yields an indeterminate form, the limit can be evaluated as:
[tex]\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \][/tex]
provided the limit on the right exists. Here:
[tex]\[ f(x) = \sqrt{2 + \sqrt{2 + x}} - 2, \quad g(x) = x - 2 \][/tex]
3. Differentiate the Numerator and Denominator:
Differentiate [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ f(x) = \sqrt{2 + \sqrt{2 + x}} - 2 \][/tex]
The derivative [tex]\( f'(x) \)[/tex] involves the chain rule. Let [tex]\( u = 2 + \sqrt{2 + x} \)[/tex]. Then,
[tex]\[ f(x) = \sqrt{u} - 2 \][/tex]
So,
[tex]\[ f'(x) = \frac{d}{dx} (\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \][/tex]
Now,
[tex]\[ \frac{du}{dx} = \frac{d}{dx} (2 + \sqrt{2 + x}) = \frac{d}{dx} (\sqrt{2 + x}) = \frac{1}{2\sqrt{2 + x}} \][/tex]
Putting it all together,
[tex]\[ f'(x) = \frac{1}{2\sqrt{2 + \sqrt{2 + x}}} \cdot \frac{1}{2\sqrt{2 + x}} \][/tex]
Simplify:
[tex]\[ f'(x) = \frac{1}{4\sqrt{2 + \sqrt{2 + x}} \sqrt{2 + x}} \][/tex]
The derivative of the denominator [tex]\( g(x) = x - 2 \)[/tex] is simply:
[tex]\[ g'(x) = 1 \][/tex]
4. Apply L'Hopital's Rule:
Now, we can apply L'Hopital's Rule:
[tex]\[ \lim_{x \to 2} \frac{\sqrt{2 + \sqrt{2 + x}} - 2}{x - 2} = \lim_{x \to 2} \frac{\frac{1}{4\sqrt{2 + \sqrt{2 + x}} \sqrt{2 + x}}}{1} \][/tex]
5. Evaluate the Limit:
Substitute [tex]\( x = 2 \)[/tex] in the simplified derivative:
[tex]\[ \lim_{x \to 2} \frac{1}{4\sqrt{2 + \sqrt{2 + x}} \sqrt{2 + x}} = \frac{1}{4\sqrt{2 + \sqrt{4}} \sqrt{4}} = \frac{1}{4\sqrt{2 + 2} \cdot 2} = \frac{1}{4 \cdot 2 \cdot 2} = \frac{1}{16} \][/tex]
6. Final Answer:
Therefore, the limit is:
[tex]\[ \lim _{x \rightarrow 2} \frac{\sqrt{2 + \sqrt{2 + x}} - 2}{x - 2} = \frac{1}{16} \][/tex]
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