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Guess the value of the limit by considering the dominant terms in the numerator and denominator. Then use I'Hospital's Rule to confirm your guess.$$\lim _{x \rightarrow \infty} \frac{e^{-2 x}+x+e^{0.1 x}}{x^{3}-x^{2}}$$

$\lim _{x \rightarrow \infty} \frac{e^{-2 x}+x+e^{0.1 x}}{x^{3}-x^{2}}=\infty$

Calculus 1 / AB

Chapter 4

Applications of Derivatives

Section 3

L'Hospital's Rule: Comparing Rates of Growth

Differentiation

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

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All right. So here we are, given the limit as X approaches infinity, uh, e to the power of negative to X plus tax plus e the part of 0.1 X over X cubed minus X squared is we want to guess what we think. That limit would be so the dominant terms in the denominator, the fastest growing one will be this x cubed. And in the numerator, the fastest growing one should be this e to the positive value of X. Even though it's a small value, these exponential functions were ese their numbers raised ex typically grow faster, not totally, always will grow faster than, ah x rays to a number. And so I would anticipate the new reader growing faster than the denominator, meaning the limit should be infinity. But But let's find out, we'll use Lope Ito's rule. Okay, so we get the limit as X approaches infinity. And so, of course, if we evaluate this, we're gonna get infinity plus infinity plus infinity over infinity minus infinity is still We end up with, uh, infinity over infinity when he violated. And so we take low Patel's ruling. We'll get the derivative of the numerator. So it's gonna be negative to e to the power of negative to X plus one plus 0.1 e to the 0.1 x Okay. And then of the denominator, we get three x squared, minus two X. So again evaluated. Uh, we get infinity plus one plus infinity over infinity minus infinity. So we need to take Luke. Patrols rule again, and this will give us limit as X approaches infinity of it's time to get four times e to the negative to X. The one becomes a zero on plus 0.1 e to the 0.1 x over six X minus two. Okay. And so again, evaluated. This is gonna be infinity over infinity. You. So we need to keep on going. And so this will give us the limit. As X approaches infinity of now, we'll get a negative eight e to the negative to x plus 0.1 e to the 0.1 x over six. Okay, so remember we can rewrite this e to the negative to X, and that's going to be the same as won over e to the two X. So when we evaluate our limit this time we get negative. 81 over Infinity plus 0.1 e times the power of 0.1 times infinity over six, which is gonna give us zero plus infinity over six. This is equal to infinity. So are initial guesses based on the dominating functions here was correct. The limit is indeed infinity.

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