At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

If [tex]\lim_{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80[/tex], find [tex]n[/tex].

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].