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Sagot :
To solve for [tex]\( n \)[/tex] in the limit problem [tex]\(\operatorname{Lt}_{x \rightarrow 2} \frac{x^n - 2^n}{x - 2} = 80\)[/tex], let's break it down step-by-step.
1. Understand the Limit Expression:
[tex]\[ \operatorname{Lt}_{x \rightarrow 2} \frac{x^n - 2^n}{x - 2} \][/tex]
As [tex]\( x \)[/tex] approaches 2, the expression [tex]\(\frac{x^n - 2^n}{x - 2}\)[/tex] resembles the difference quotient, which is often used to find derivatives.
2. Simplify Using L'Hôpital's Rule:
Since directly substituting [tex]\( x = 2 \)[/tex] results in the form [tex]\(\frac{0}{0}\)[/tex], we can apply L'Hôpital's rule. This rule states that:
[tex]\[ \operatorname{Lt}_{x \rightarrow c} \frac{f(x)}{g(x)} = \operatorname{Lt}_{x \rightarrow c} \frac{f'(x)}{g'(x)} \quad \text{if the limit is of the form} \quad \frac{0}{0} \text{ or } \frac{\infty}{\infty} \][/tex]
In our case:
[tex]\[ f(x) = x^n - 2^n \quad \text{and} \quad g(x) = x - 2 \][/tex]
3. Differentiate the Numerator and Denominator:
[tex]\[ f'(x) = \frac{d}{dx} (x^n - 2^n) = nx^{n-1} \][/tex]
(Note that [tex]\( 2^n \)[/tex] is a constant with respect to [tex]\( x \)[/tex], so its derivative is 0)
[tex]\[ g'(x) = \frac{d}{dx} (x - 2) = 1 \][/tex]
4. Apply L'Hôpital's Rule:
[tex]\[ \operatorname{Lt}_{x \rightarrow 2} \frac{x^n - 2^n}{x - 2} = \operatorname{Lt}_{x \rightarrow 2} \frac{nx^{n-1}}{1} = n \cdot 2^{n-1} \][/tex]
5. Set the Limit Equal to 80:
According to the given condition:
[tex]\[ n \cdot 2^{n-1} = 80 \][/tex]
6. Solve for [tex]\( n \)[/tex]:
Let's find an integer [tex]\( n \)[/tex] that satisfies this equation:
[tex]\[ n \cdot 2^{n-1} = 80 \][/tex]
Try different integer values for [tex]\( n \)[/tex]:
For [tex]\( n = 4 \)[/tex]:
[tex]\[ 4 \cdot 2^{4-1} = 4 \cdot 2^3 = 4 \cdot 8 = 32 \quad (\text{Not } 80) \][/tex]
For [tex]\( n = 5 \)[/tex]:
[tex]\[ 5 \cdot 2^{5-1} = 5 \cdot 2^4 = 5 \cdot 16 = 80 \quad (\text{Correct}) \][/tex]
Therefore, [tex]\( n = 5 \)[/tex] is the correct value.
7. Conclusion:
[tex]\[ n = 5 \][/tex]
Thus, the value of [tex]\( n \)[/tex] that satisfies the given limit equation is [tex]\( \boxed{5} \)[/tex].
1. Understand the Limit Expression:
[tex]\[ \operatorname{Lt}_{x \rightarrow 2} \frac{x^n - 2^n}{x - 2} \][/tex]
As [tex]\( x \)[/tex] approaches 2, the expression [tex]\(\frac{x^n - 2^n}{x - 2}\)[/tex] resembles the difference quotient, which is often used to find derivatives.
2. Simplify Using L'Hôpital's Rule:
Since directly substituting [tex]\( x = 2 \)[/tex] results in the form [tex]\(\frac{0}{0}\)[/tex], we can apply L'Hôpital's rule. This rule states that:
[tex]\[ \operatorname{Lt}_{x \rightarrow c} \frac{f(x)}{g(x)} = \operatorname{Lt}_{x \rightarrow c} \frac{f'(x)}{g'(x)} \quad \text{if the limit is of the form} \quad \frac{0}{0} \text{ or } \frac{\infty}{\infty} \][/tex]
In our case:
[tex]\[ f(x) = x^n - 2^n \quad \text{and} \quad g(x) = x - 2 \][/tex]
3. Differentiate the Numerator and Denominator:
[tex]\[ f'(x) = \frac{d}{dx} (x^n - 2^n) = nx^{n-1} \][/tex]
(Note that [tex]\( 2^n \)[/tex] is a constant with respect to [tex]\( x \)[/tex], so its derivative is 0)
[tex]\[ g'(x) = \frac{d}{dx} (x - 2) = 1 \][/tex]
4. Apply L'Hôpital's Rule:
[tex]\[ \operatorname{Lt}_{x \rightarrow 2} \frac{x^n - 2^n}{x - 2} = \operatorname{Lt}_{x \rightarrow 2} \frac{nx^{n-1}}{1} = n \cdot 2^{n-1} \][/tex]
5. Set the Limit Equal to 80:
According to the given condition:
[tex]\[ n \cdot 2^{n-1} = 80 \][/tex]
6. Solve for [tex]\( n \)[/tex]:
Let's find an integer [tex]\( n \)[/tex] that satisfies this equation:
[tex]\[ n \cdot 2^{n-1} = 80 \][/tex]
Try different integer values for [tex]\( n \)[/tex]:
For [tex]\( n = 4 \)[/tex]:
[tex]\[ 4 \cdot 2^{4-1} = 4 \cdot 2^3 = 4 \cdot 8 = 32 \quad (\text{Not } 80) \][/tex]
For [tex]\( n = 5 \)[/tex]:
[tex]\[ 5 \cdot 2^{5-1} = 5 \cdot 2^4 = 5 \cdot 16 = 80 \quad (\text{Correct}) \][/tex]
Therefore, [tex]\( n = 5 \)[/tex] is the correct value.
7. Conclusion:
[tex]\[ n = 5 \][/tex]
Thus, the value of [tex]\( n \)[/tex] that satisfies the given limit equation is [tex]\( \boxed{5} \)[/tex].
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