Can L'hopital's Rule be Applied to this Limit?

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    l'hopital Limit
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Discussion Overview

The discussion revolves around the application of L'Hôpital's Rule to a limit problem involving trigonometric functions as \( x \) approaches infinity. Participants explore the formulation of the limit and the steps necessary to analyze it, particularly focusing on the logarithmic transformation of the expression.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant presents a limit expression involving \( \tan \) and \( \sin \) as \( x \) approaches infinity.
  • Another participant points out a typographical error in the limit expression and seeks clarification on its correct form.
  • A later reply suggests rewriting the limit in terms of \( y \) and taking the natural logarithm to facilitate the application of L'Hôpital's Rule, indicating that this transformation leads to a \( \frac{0}{0} \) indeterminate form.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the application of L'Hôpital's Rule, as the discussion includes corrections and suggestions without a definitive resolution on the limit itself.

Contextual Notes

The discussion includes potential assumptions about the behavior of the trigonometric functions as \( x \) approaches infinity, which may not be fully articulated. The transformation steps leading to the \( \frac{0}{0} \) form are also not exhaustively detailed.

hadi amiri 4
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[tex]\lim_{x \rightarrow infinity}/ left(\frac{\1+Tan\frac{/pi}{2x}}{1+sin\frac{/pi}{3x}}}right)^x[/tex]
 
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sorry i made mistake in typing
 


So what does the limit problem actually look like?
 


hadi amiri 4 said:
[tex]\lim_{x \rightarrow infinity}/ left(\frac{\1+Tan\frac{/pi}{2x}}{1+sin\frac{/pi}{3x}}}right)^x[/tex]

I assume you meant

[tex]\lim_{x \rightarrow {\infty}} \left(\frac{1 \, + \, tan\left(\frac{\pi}{2x}\right)}{1 \, + \, sin \left(\frac{\pi}{3x}\right)} \right)^{x}[/tex]

then I would suggest letting that = y, then take ln of both sides and you should get something like:

[tex]ln(y) \, = \, \lim_{x \rightarrow {\infty}} \frac{ln\left(\frac{1 \, + \, tan\left(\frac{\pi}{2x}\right)}{1 \, + \, sin \left(\frac{\pi}{3x}\right)} \right)}{\frac{1}{x}}[/tex]

which if you "plug in" the limit should give you [tex]\frac{0}{0}[/tex] making it a candidate for L'hopital. Try that.
 

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