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Sagot :
To solve the problem, we need to determine the limit of the given expression as [tex]\( x \)[/tex] approaches 2 and then solve for the positive integer [tex]\( n \)[/tex] such that this limit equals 32.
1. Expression Analysis:
We are given:
[tex]\[ \lim_{x \to 2} \frac{x^n - 2^n}{x - 2} \][/tex]
This is a classical limit form that suggests we might use L'Hôpital's Rule or polynomial properties. However, there's a more straightforward polynomial approach for such limits.
2. Form of the Limit:
Notice that [tex]\( x^n - 2^n \)[/tex] can be factored as:
[tex]\[ x^n - 2^n = (x - 2) \cdot Q(x) \][/tex]
where [tex]\( Q(x) \)[/tex] is a polynomial. Specifically, for [tex]\( x \to 2 \)[/tex], [tex]\( Q(x) \)[/tex] encapsulates terms derived from the binomial theorem. The critical insight is that as [tex]\( x \)[/tex] approaches 2, [tex]\( Q(x) \)[/tex] approaches a specific value.
3. Evaluating the Limit:
When [tex]\( x \)[/tex] is very close to 2, we can simplify:
[tex]\[ \frac{x^n - 2^n}{x - 2} \approx Q(2) \][/tex]
4. Finding the Polynomial [tex]\( Q(x) \)[/tex]:
From polynomial properties and knowing the derivative's role in such context, it can be deduced:
[tex]\[ Q(2) = 2^{n-1} \cdot n \][/tex]
5. Given the Limit Value:
We are given:
[tex]\[ \lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = 32 \][/tex]
Therefore:
[tex]\[ 2^{n-1} \cdot n = 32 \][/tex]
6. Solving for [tex]\( n \)[/tex]:
To find [tex]\( n \)[/tex], solve the equation:
[tex]\[ 2^{n-1} \cdot n = 32 \][/tex]
We know [tex]\( 32 = 2^5 \)[/tex], so rewrite the equation:
[tex]\[ 2^{n-1} \cdot n = 2^5 \][/tex]
Equate the exponents of 2:
[tex]\[ 2^{n - 1} \cdot n = 2^5 \][/tex]
So [tex]\( 2^{n-1} \)[/tex] must equal 2 raised to a power near 5 divided by [tex]\( n \)[/tex]:
[tex]\[ 2^{n-1} = 2^5 \div n \][/tex]
We hypothesize various values of [tex]\( n \)[/tex]. Let's check if [tex]\( n = 4 \)[/tex] satisfies the equation:
[tex]\[ 2^{4-1} \cdot 4 = 2^3 \cdot 4 = 8 \cdot 4 = 32 \][/tex]
7. Conclusion:
Hence, the positive integer [tex]\( n \)[/tex] that satisfies the given limit condition is [tex]\( n = 4 \)[/tex]. Therefore, the solution is:
[tex]\[ n = 4 \][/tex]
1. Expression Analysis:
We are given:
[tex]\[ \lim_{x \to 2} \frac{x^n - 2^n}{x - 2} \][/tex]
This is a classical limit form that suggests we might use L'Hôpital's Rule or polynomial properties. However, there's a more straightforward polynomial approach for such limits.
2. Form of the Limit:
Notice that [tex]\( x^n - 2^n \)[/tex] can be factored as:
[tex]\[ x^n - 2^n = (x - 2) \cdot Q(x) \][/tex]
where [tex]\( Q(x) \)[/tex] is a polynomial. Specifically, for [tex]\( x \to 2 \)[/tex], [tex]\( Q(x) \)[/tex] encapsulates terms derived from the binomial theorem. The critical insight is that as [tex]\( x \)[/tex] approaches 2, [tex]\( Q(x) \)[/tex] approaches a specific value.
3. Evaluating the Limit:
When [tex]\( x \)[/tex] is very close to 2, we can simplify:
[tex]\[ \frac{x^n - 2^n}{x - 2} \approx Q(2) \][/tex]
4. Finding the Polynomial [tex]\( Q(x) \)[/tex]:
From polynomial properties and knowing the derivative's role in such context, it can be deduced:
[tex]\[ Q(2) = 2^{n-1} \cdot n \][/tex]
5. Given the Limit Value:
We are given:
[tex]\[ \lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = 32 \][/tex]
Therefore:
[tex]\[ 2^{n-1} \cdot n = 32 \][/tex]
6. Solving for [tex]\( n \)[/tex]:
To find [tex]\( n \)[/tex], solve the equation:
[tex]\[ 2^{n-1} \cdot n = 32 \][/tex]
We know [tex]\( 32 = 2^5 \)[/tex], so rewrite the equation:
[tex]\[ 2^{n-1} \cdot n = 2^5 \][/tex]
Equate the exponents of 2:
[tex]\[ 2^{n - 1} \cdot n = 2^5 \][/tex]
So [tex]\( 2^{n-1} \)[/tex] must equal 2 raised to a power near 5 divided by [tex]\( n \)[/tex]:
[tex]\[ 2^{n-1} = 2^5 \div n \][/tex]
We hypothesize various values of [tex]\( n \)[/tex]. Let's check if [tex]\( n = 4 \)[/tex] satisfies the equation:
[tex]\[ 2^{4-1} \cdot 4 = 2^3 \cdot 4 = 8 \cdot 4 = 32 \][/tex]
7. Conclusion:
Hence, the positive integer [tex]\( n \)[/tex] that satisfies the given limit condition is [tex]\( n = 4 \)[/tex]. Therefore, the solution is:
[tex]\[ n = 4 \][/tex]
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