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I am trying to evaluate the following limit:

$$ \lim_{t \to 0} \frac{\sin(t) - \cos(t) + e^t}{t^3}. $$

I would like to avoid using L'Hôpital's Rule or Taylor series expansions.

Is there a more "direct" or classical method to approach this limit? Any detailed steps or insights would be greatly appreciated.

I have this feeling that I might inspire from a previous limit but I am really stuck.

CHOSM
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    Are you sure the numerator is correct? The terms of lower order than $t^3$ in its Maclaurin series don't cancel. They would, for example, in $\sin t-\cosh t+e^{-t}$. – J.G. Dec 07 '24 at 19:59
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    Hint: add and subtract $1$ in the numerator and use three well known limits – Davide Masi Dec 07 '24 at 20:04
  • @DavideMasi The problem is with the second limit it'll be $\pm\infty$ and the others will be $+\infty$ thus an undetermined form! – CHOSM Dec 07 '24 at 20:11
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    Why would you like to avoid Taylor's expansion? It is like asking a mammal to survive without its heart. – mathcounterexamples.net Dec 07 '24 at 20:15
  • @CHOSM I don't understand what you mean. You should write $\frac{\sin(t) - \cos(t) + e^t}{t^3}$ as $\frac{1}{t^2}\frac{\sin(t) - \cos(t) + e^t}{t}$ – Davide Masi Dec 07 '24 at 20:16
  • @DavideMasi it's the $\frac{1-cost}{t^2}$ that is causing the problem. Did I think of the wrong limit? – CHOSM Dec 07 '24 at 20:19

1 Answers1

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Observe that $\frac{\sin(t) - \cos(t) + e^t}{t^3}$ = $\frac{1}{t^2}\frac{\sin(t) - \cos(t) + e^t}{t}$ = $\frac{1}{t^2}\frac{\sin(t) + [1- \cos(t)] + [e^t - 1]}{t}$ = $\frac{1}{t^2}\left[\frac{\sin(t)}{t} + \frac{1 - \cos(t)}{t} + \frac{e^t-1}{t}\right].$

You should know that the limits of the fractions are respectively $1,0, 1$. It is easy to conclude from here!

Davide Masi
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