Evaluating Limits using L'Hospital

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Discussion Overview

The discussion revolves around evaluating the limit of the expression $$\frac{e^{2t} - 1}{1 - \cos(t)}$$ as \( t \) approaches 0, using L'Hôpital's Rule. Participants explore the application of derivatives and the implications of one-sided limits in determining the existence of the limit.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions their work on the limit, noting they obtained an undefined result after taking derivatives twice.
  • Another participant suggests that the limit does not exist after one application of L'Hôpital's Rule and encourages examining one-sided limits.
  • A participant expresses confusion about the implications of the cosine function in the denominator and seeks clarification on whether the limit should be labeled as "does not exist" or "undefined."
  • A later reply provides a detailed calculation, confirming the limit is an indeterminate form and shows that the one-sided limits approach negative and positive infinity, leading to the conclusion that the limit does not exist.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the terminology for the limit's status, with some suggesting "does not exist" and others considering "undefined." The discussion reflects multiple viewpoints on the interpretation of the limit's behavior.

Contextual Notes

The discussion includes assumptions about the application of L'Hôpital's Rule and the behavior of trigonometric functions near zero, which may not be fully explored or agreed upon.

shamieh
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Need someone to check my work.lim t -> 0

$$\frac{e^{2t} - 1}{1 - cos(t)}$$

after I took the derivative twice

I got $$\frac{2}{0}$$ = undefined?
 
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You can determine the limit does not exist after just one application of L'Hôpital's Rule. I would also look at the one-sided limits. What to you find?

Note: use the $\LaTeX$ code \lim_{t\to0} for your limit.
 
Oh because the denominator has the cos and sin. that's what you are saying correct? How I could of known after 1 application of LHopsital...So is this the correct answer though? Does not exist? Or Should I put "Undefined"?
 
This is what I would write:

$$L=\lim_{t\to0}\frac{e^{2t}-1}{1-\cos(t)}$$

This is the indeterminate form $$\frac{0}{0}$$, so application of L'Hôpital's rule yields:

$$L=\lim_{t\to0}\frac{2e^{2t}}{\sin(t)}=\frac{2}{0}$$

This is undefined, so we want to look at the one-sided limits:

$$\lim_{t\to0^{-}}\frac{2e^{2t}}{\sin(t)}=-\infty$$

$$\lim_{t\to0^{+}}\frac{2e^{2t}}{\sin(t)}=\infty$$

Since:

$$\lim_{t\to0^{-}}\frac{2e^{2t}}{\sin(t)}\ne \lim_{t\to0^{+}}\frac{2e^{2t}}{\sin(t)}$$ we may conclude that the limit $L$ does not exist.
 

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