MHB When taking the limit at infinity, is this allowed?

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The discussion centers on the limit at infinity, specifically the expression $$\lim_{t \to \infty}\frac{te^{-st}}{-s}$$ being evaluated as 0. Participants question the reasoning that $$e^t$$ grows faster than $$t$$ and whether it is valid to treat infinity in this manner. The concept of the indeterminate form $$\frac{\infty}{\infty}$$ is emphasized, suggesting the application of L'Hôpital's Rule to resolve such limits. The mention of L'Hôpital's Rule indicates a method for handling limits involving infinity. Overall, the conversation highlights the complexities of evaluating limits at infinity and the importance of understanding indeterminate forms.
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In this video of Laplace transforms the equation $$\lim_{t \to \infty}\frac{te^{-st}}{-s}$$ is said to be 0. I'm not sure I agree with the reasoning. It says it's because $$e^t$$ grows faster than $$t$$; can you treat infinity like that? For example could you say $$\lim_{x \to \infty}\frac{x}{x^2}=0$$? I thought it was undefined because obviously $$\frac{infty}{infty}$$ is undefined.
 
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The form $$\frac{\infty}{\infty}$$ is indeterminate...to allay your concerns, apply L'Hôpital's Rule. :D
 
MarkFL said:
The form $$\frac{\infty}{\infty}$$ is indeterminate...to allay your concerns, apply L'Hôpital's Rule. :D

Maybe the video was aluding to L'Hôpital's Rule when it said $$e^t$$ grows faster than $$t$$.

By the way, according to here $$\frac{\infty}{\infty}$$ is indeterminate.
 

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