Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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].
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
