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\(5-64\) Find the limit. Use I'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow 1} \frac{x^{9}-1}{x^{5}-1}$$

Short Answer

Expert verified
The limit is \( \frac{9}{5} \).

Step by step solution

01

Determine the form of the limit

First, substitute the value of \(x\) approaching 1 into both the numerator and the denominator of the function \( \frac{x^9-1}{x^5-1} \). Evaluating gives \( x^9 - 1 = 1^9 - 1 = 0 \) and \( x^5 - 1 = 1^5 - 1 = 0 \).Since both the numerator and the denominator equal 0, this is an indeterminate form of \( \frac{0}{0} \). So, we can apply L'Hospital's Rule.
02

Apply L'Hospital's Rule

L'Hospital's Rule states that if the limit results in an indeterminate form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]where \( f'(x) \) and \( g'(x) \) are the derivatives of \( f(x) \) and \( g(x) \) respectively.Find the derivatives: - Derivative of the numerator \( f(x) = x^9 - 1 \) is \( f'(x) = 9x^8 \).- Derivative of the denominator \( g(x) = x^5 - 1 \) is \( g'(x) = 5x^4 \).
03

Evaluate the limit with calculated derivatives

Now, apply the derivatives to evaluate the new limit:\[ \lim_{x \to 1} \frac{9x^8}{5x^4} = \lim_{x \to 1} \frac{9x^4}{5} \]Substitute \( x = 1 \):\[ \frac{9(1)^4}{5} = \frac{9}{5} \]
04

Conclusion

Since the limit \( \lim_{x \to 1} \frac{x^9-1}{x^5-1} \) simplifies to \( \frac{9}{5} \), this is the limit of the original function as \( x \) approaches 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L'Hospital's Rule
L'Hospital's Rule is a powerful tool in calculus. It helps us evaluate limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In these cases, direct substitution doesn't give useful results, so we use calculus to find a solution. Here's how L'Hospital's Rule works: if you have a limit of the form \( \lim_{x \to a} \frac{f(x)}{g(x)} \) resulting in an indeterminate form, you can replace it with \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \).
  • This means you take the derivative of the numerator and the derivative of the denominator.
  • Then, re-evaluate the limit using these derivatives instead of the original functions.

L'Hospital's Rule often simplifies the problem to a form where direct substitution is again possible. This method can be repeated if the result still remains an indeterminate form, although usually, it only takes one application to resolve the problem.
Limit Evaluation
Limit evaluation is a fundamental process in calculus. It involves determining the value a function approaches as the input gets closer to a certain point. Calculating limits is essential for understanding behavior at boundaries, like points of discontinuity or asymptotes.
When you evaluate a limit, you typically start by directly substituting the point into the function:
  • If the result is a real number, that's the limit.
  • If a form like \( \frac{0}{0} \) arises, you may need additional tools like L'Hospital's Rule.

Sometimes, functions can have simpler alternate forms that are easier to evaluate. This can involve algebraic manipulation, simplifying the expression, or using known limit rules. Identifying the best approach is vital in finding the correct limit efficiently.
Indeterminate Forms
In calculus, indeterminate forms occur when evaluating a limit results in an undefined or ambiguous expression. Examples of indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( 1^\infty \), and \( \infty^0 \).
Encountering an indeterminate form indicates that further analysis or manipulation is necessary. Techniques like L'Hospital's Rule, algebraic rearrangement, or logarithmic differentiation come into play:
  • These methods help provide clarity in evaluating the ambiguous result.
  • A clear understanding of these strategies reduces the complexity, making calculus problems more approachable.

Recognizing indeterminate forms is a crucial skill, as it signals the need for advanced calculus tools to find meaningful answers in mathematical analysis.

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Most popular questions from this chapter

Ornithologists have determined that some species of birds tend to avoid flights over large bodies of water during daylight hours. It is believed that more energy is required to fly over water than over land because air generally rises over land and falls over water during the day. A bird with these tendencies is released from an island that is 5\(\mathrm { km }\) from the nearest point \(B\) on a straight shoreline, flies to a point \(C\) on the shoreline, and then flies along the shoreline to its nesting area \(D .\) Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points \(B\) and \(D\) are 13\(\mathrm { km }\) apart. (a) In general, if it takes 1.4 times as much energy to fly over water as it does over land, to what point \(C\) should the bird fly in order to minimize the total energy expended in returning to its nesting area? (b) Let \(W\) and \(L\) denote the energy (in joules) per kilometer flown over water and land, respectively. What would a large value of the ratio \(W / L\) mean in terms of the bird's flight? What would a small value mean? Determine the ratio \(W / L\) corresponding to the minimum expenditure of energy. (c) What should the value of \(W / L\) be in order for the bird to fly directly to its nesting area \(D ?\) What should the value of \(W / L\) be for the bird to fly to \(B\) and then along the shore to \(D ?\) (d) If the omithologists observe that birds of a certain species reach the shore at a point 4\(\mathrm { km }\) from \(B\) , how many times more energy does it take a bird to fly over water than over land?

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