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

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    l'hopital Limit
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SUMMARY

The limit problem presented is \lim_{x \rightarrow \infty} \left(\frac{1 + \tan\left(\frac{\pi}{2x}\right)}{1 + \sin\left(\frac{\pi}{3x}\right)}\right)^{x}. To solve this limit, it is recommended to set the expression equal to y and take the natural logarithm of both sides, resulting in 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}}. This transformation leads to a 0/0 indeterminate form, making it suitable for application of L'Hôpital's Rule.

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


So what does the limit problem actually look like?
 


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

I assume you meant

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

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

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}}

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

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