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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_{x\to \infty} \left( \frac{2x - 3}{2x + 5} \right)^{2x + 1} $

$e^{-8}$

04:43

Wen Z.

01:49

Carson M.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Derivatives

Differentiation

Volume

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So this problem we're going to be taking the natural log of both sides to start. And that will allow us to put it in a simpler form. That's going to be the natural log of two X -3. Um Natural log of two x -3 Over two X Plus five. And that is all gonna be over one divided by two x plus one. When we plug in X equals infinity. We get 0/0 which is the indeterminate form. Um So we're gonna use local towels role. Then using that we're able to end up getting negative uh negative eight times choo X Over X. Class one over X. Weird. And then on the denominator were able to get choo X over X Plus five over acts times two. X. Overact -3 over X. And the reason we do this is so that way when we evaluate it to infinity a lot of these things go to zero. Just leaving us with a -8. So since this is the natural log of Y equals negative eight. Our final answer is going to be Y equals E. To the negative age

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