Finding Limits Approaching Infinity for Improper Integrals

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SUMMARY

The discussion focuses on evaluating the limit of the expression \( \lim_{t \to -\infty} \left(e^{2} - \frac{t-3}{3} e^{\frac{t}{3}}\right) \) to determine the convergence or divergence of an improper integral. Participants clarify that as \( t \) approaches negative infinity, the term \( e^{\frac{t}{3}} \) approaches zero, leading to the conclusion that the limit converges to \( e^{2} \). The discussion emphasizes the importance of understanding horizontal asymptotes and the behavior of exponential functions in limit evaluations.

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  • Understanding of limits in calculus
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  • Knowledge of horizontal asymptotes
  • Basic proficiency in exponential functions
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Homework Statement



Lim ((e^{2})-(\frac{t-3}{3}( * e^^{\frac{t}{3}})))
t -> -\infty

[sorry for the formatting, I tried my best! that is "The limit as t approaches negative infinity of e squared minus (t-3/3) e to the t/3)]"]

Homework Equations


I am solving improper integrals to find out if this is convergent or divergent and do not remember how to show on paper to solve for a limit.

The Attempt at a Solution


Well, if memory serves, the limit as we approach negative infinity is a horizontal asymptote, right? But I only remember those silly little coefficient tricks to find a horizontal asymptote. Otherwise, I'd just look on a calculator. Someone tried to explain it to me and said I should think of it as going towards positive infinity first and then negate it ... but even with positive infinity I don't no where to start!

Your detailed instructions/explanations would be awesome! I really want to understand how to do this by hand.
 
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\left(e^2- \frac{t-3}{3}\right)e^{-t/3}
or
e^2- \frac{t-3}{3}e^{-t/3}?

In either case it looks pretty straight forward to me: e-x, times any power of t, goes to 0 as x goes to infinity.
 

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