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Find the limit. Use l'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.

$ \displaystyle \lim_{t\to 1} \frac{t^8 - 1}{t^5 - 1} $

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This limit has the form$\frac{0}{0} \cdot \lim _{t \rightarrow 1} \frac{t^{8}-1}{t^{5}-1} \stackrel{\mathrm{H}}{=} \lim _{t \rightarrow 1} \frac{8 t^{7}}{5 t^{4}}=\frac{8}{5} \lim _{t \rightarrow 1} t^{3}=\frac{8}{5}(1)=\frac{8}{5}$

02:33

Mengsha Yao

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Derivatives

Differentiation

Volume

Campbell University

Harvey Mudd College

University of Nottingham

Idaho State University

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so there is a way to do this without looked as role. It would require a lot of factoring now. Um and this is not preferred. So something there's two kind of options that we can use. Um One is a little bit more insightful ones, a little bit more conceptual and then the other ones using towels role. So using local tells rule, we see that we would have T the eight and it's one over T to the five minus one. What we would see happening because we would have a T To the 7th over five T to the fourth. We would keep doing this because this would go on for a while and then we keep plugging in. Um The limit as T equals one. However, something else that we could do is if we do this and use low battles rule, we can also factor that would be another option. And when we factor we see that We're able to cancel two terms and then cancelling two terms. We can evaluate the limit at T equals one And that's going to end up giving us 8/5 as a result for the final answer

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