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Sagot :
Certainly! Let's solve the limit:
[tex]\[ \lim_{x \rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} \][/tex]
### Step-by-Step Solution
#### Step 1: Simplify the Limit Expression in the Numerator
First, observe the behavior of the function in the numerator as [tex]\( x \)[/tex] approaches 0:
- [tex]\( 27^x, 9^x, \)[/tex] and [tex]\( 3^x \)[/tex] are all exponential functions with positive bases.
- Specifically, [tex]\( 27 = 3^3 \)[/tex], [tex]\( 9 = 3^2 \)[/tex].
We can rewrite [tex]\( 27^x \)[/tex] and [tex]\( 9^x \)[/tex] using the base 3:
- [tex]\( 27^x = (3^3)^x = 3^{3x} \)[/tex],
- [tex]\( 9^x = (3^2)^x = 3^{2x} \)[/tex].
This simplifies the numerator to:
[tex]\[ 27^x - 9^x - 3^x + 1 = 3^{3x} - 3^{2x} - 3^x + 1 \][/tex]
As [tex]\( x \rightarrow 0 \)[/tex]:
- [tex]\( 3^{3x} \rightarrow 1 \)[/tex],
- [tex]\( 3^{2x} \rightarrow 1 \)[/tex],
- [tex]\( 3^x \rightarrow 1 \)[/tex].
Subsequently, the expressions [tex]\( 27^x - 9^x \)[/tex] and [tex]\( 3^x - 1 \)[/tex] approach 0. However, for precise behavior, we need to incorporate the corrections using Taylor expansions.
Taylor expansion for [tex]\( e^{f(x)} \approx 1 + f(x) \)[/tex] for small [tex]\( f(x) \)[/tex]:
- [tex]\( 3^{kx} \approx 1 + kx \ln 3 \)[/tex].
Thus,
- [tex]\( 3^{3x} \approx 1 + 3x \ln 3 \)[/tex],
- [tex]\( 3^{2x} \approx 1 + 2x \ln 3 \)[/tex],
- [tex]\( 3^x \approx 1 + x \ln 3 \)[/tex].
Plugging these into [tex]\( 3^{3x} - 3^{2x} - 3^x + 1 \)[/tex]:
[tex]\[ (1 + 3x \ln 3) - (1 + 2x \ln 3) - (1 + x \ln 3) + 1 = 3x \ln 3 - 2x \ln 3 - x \ln 3 = 0 \][/tex]
#### Step 2: Simplify the Denominator
For the denominator [tex]\( \sqrt{2} - \sqrt{1 + \cos x} \)[/tex], as [tex]\( x \rightarrow 0 \)[/tex]:
- [tex]\( \cos x \approx 1 - \frac{x^2}{2} \)[/tex], using the Taylor expansion for cosine around 0.
Therefore,
[tex]\[ \sqrt{1 + \cos x} \approx \sqrt{1 + 1 - \frac{x^2}{2}} = \sqrt{2 - \frac{x^2}{2}} \][/tex]
Using [tex]\( \sqrt{a + b} \approx \sqrt{a} (1 + \frac{b}{2a}) \)[/tex] for small [tex]\( b \)[/tex],
[tex]\[ \sqrt{2 - \frac{x^2}{2}} \approx \sqrt{2}\left( 1 - \frac{x^2}{8} \right) \][/tex]
So,
[tex]\[ \sqrt{2} - \sqrt{1 + \cos x} \approx \sqrt{2} - \sqrt{2} \left( 1 - \frac{x^2}{8} \right) = \sqrt{2} \cdot \frac{x^2}{8} = \frac{\sqrt{2} x^2}{8} \][/tex]
#### Step 3: Form the Limit
Putting it all together, the limit is
[tex]\[ \lim_{x \rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = \lim_{x \rightarrow 0} \frac{0}{\frac{\sqrt{2} x^2}{8}} \][/tex]
For the behavior near 0 using terms:
- The numerator [tex]\( 3^{3x} - 3^{2x} - 3^x + 1 \approx 3x (\ln 3)^2 \)[/tex].
So the problem reduces to the limit of:
[tex]\[ \lim_{x \rightarrow 0} \frac{3x (\ln 3)^3}{\frac{\sqrt{2} x^2}{8}} = \lim_{x \rightarrow 0} \frac{8 \times 3 (\ln 3)^2}{\sqrt{2} x} = \frac{24 (\ln 3)^2}{\sqrt{2} x} \text{ as } x \to 0 \][/tex]
Factoring, dividing out x terms:
[tex]\[ 8\sqrt{2} (\ln 3)^2 = 8 \sqrt{2} (\ln 3)^2. \][/tex]
Thus,
[tex]\[ \lim_{x \rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = 8\sqrt{2} \ln 3^2 \][/tex]
[tex]\[ \lim_{x \rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} \][/tex]
### Step-by-Step Solution
#### Step 1: Simplify the Limit Expression in the Numerator
First, observe the behavior of the function in the numerator as [tex]\( x \)[/tex] approaches 0:
- [tex]\( 27^x, 9^x, \)[/tex] and [tex]\( 3^x \)[/tex] are all exponential functions with positive bases.
- Specifically, [tex]\( 27 = 3^3 \)[/tex], [tex]\( 9 = 3^2 \)[/tex].
We can rewrite [tex]\( 27^x \)[/tex] and [tex]\( 9^x \)[/tex] using the base 3:
- [tex]\( 27^x = (3^3)^x = 3^{3x} \)[/tex],
- [tex]\( 9^x = (3^2)^x = 3^{2x} \)[/tex].
This simplifies the numerator to:
[tex]\[ 27^x - 9^x - 3^x + 1 = 3^{3x} - 3^{2x} - 3^x + 1 \][/tex]
As [tex]\( x \rightarrow 0 \)[/tex]:
- [tex]\( 3^{3x} \rightarrow 1 \)[/tex],
- [tex]\( 3^{2x} \rightarrow 1 \)[/tex],
- [tex]\( 3^x \rightarrow 1 \)[/tex].
Subsequently, the expressions [tex]\( 27^x - 9^x \)[/tex] and [tex]\( 3^x - 1 \)[/tex] approach 0. However, for precise behavior, we need to incorporate the corrections using Taylor expansions.
Taylor expansion for [tex]\( e^{f(x)} \approx 1 + f(x) \)[/tex] for small [tex]\( f(x) \)[/tex]:
- [tex]\( 3^{kx} \approx 1 + kx \ln 3 \)[/tex].
Thus,
- [tex]\( 3^{3x} \approx 1 + 3x \ln 3 \)[/tex],
- [tex]\( 3^{2x} \approx 1 + 2x \ln 3 \)[/tex],
- [tex]\( 3^x \approx 1 + x \ln 3 \)[/tex].
Plugging these into [tex]\( 3^{3x} - 3^{2x} - 3^x + 1 \)[/tex]:
[tex]\[ (1 + 3x \ln 3) - (1 + 2x \ln 3) - (1 + x \ln 3) + 1 = 3x \ln 3 - 2x \ln 3 - x \ln 3 = 0 \][/tex]
#### Step 2: Simplify the Denominator
For the denominator [tex]\( \sqrt{2} - \sqrt{1 + \cos x} \)[/tex], as [tex]\( x \rightarrow 0 \)[/tex]:
- [tex]\( \cos x \approx 1 - \frac{x^2}{2} \)[/tex], using the Taylor expansion for cosine around 0.
Therefore,
[tex]\[ \sqrt{1 + \cos x} \approx \sqrt{1 + 1 - \frac{x^2}{2}} = \sqrt{2 - \frac{x^2}{2}} \][/tex]
Using [tex]\( \sqrt{a + b} \approx \sqrt{a} (1 + \frac{b}{2a}) \)[/tex] for small [tex]\( b \)[/tex],
[tex]\[ \sqrt{2 - \frac{x^2}{2}} \approx \sqrt{2}\left( 1 - \frac{x^2}{8} \right) \][/tex]
So,
[tex]\[ \sqrt{2} - \sqrt{1 + \cos x} \approx \sqrt{2} - \sqrt{2} \left( 1 - \frac{x^2}{8} \right) = \sqrt{2} \cdot \frac{x^2}{8} = \frac{\sqrt{2} x^2}{8} \][/tex]
#### Step 3: Form the Limit
Putting it all together, the limit is
[tex]\[ \lim_{x \rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = \lim_{x \rightarrow 0} \frac{0}{\frac{\sqrt{2} x^2}{8}} \][/tex]
For the behavior near 0 using terms:
- The numerator [tex]\( 3^{3x} - 3^{2x} - 3^x + 1 \approx 3x (\ln 3)^2 \)[/tex].
So the problem reduces to the limit of:
[tex]\[ \lim_{x \rightarrow 0} \frac{3x (\ln 3)^3}{\frac{\sqrt{2} x^2}{8}} = \lim_{x \rightarrow 0} \frac{8 \times 3 (\ln 3)^2}{\sqrt{2} x} = \frac{24 (\ln 3)^2}{\sqrt{2} x} \text{ as } x \to 0 \][/tex]
Factoring, dividing out x terms:
[tex]\[ 8\sqrt{2} (\ln 3)^2 = 8 \sqrt{2} (\ln 3)^2. \][/tex]
Thus,
[tex]\[ \lim_{x \rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = 8\sqrt{2} \ln 3^2 \][/tex]
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