Is There an Error in My L'Hopital's Rule Calculation?

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    L'hopital's rule
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SUMMARY

The discussion centers on the application of L'Hôpital's Rule to evaluate the limit of the expression lim x→∞ (1 + cos(1/x))^x. The initial misunderstanding involved incorrectly identifying the limit type as 1^∞ instead of recognizing the need to apply logarithmic transformation. The correct approach involves taking the natural logarithm, leading to the limit ln|y| = x * ln(1 + cos(1/x)), which simplifies to ln(2)/0, indicating an indeterminate form that requires further analysis.

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mk2munky
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I had a test today and there was a LR limit on it that I didn't get. I thought he said he got them out of the book, but I didn't see it anywhere. The equation was:

lim x[tex]\rightarrow[/tex][tex]\infty[/tex] of (1 + cosx(1/x))x

So, I said that lim x[tex]\rightarrow[/tex][tex]\infty[/tex] of cos(1/x) = cos(0) = 1

And therefore:

lim x[tex]\rightarrow[/tex][tex]\infty[/tex] of (1 + 1)[tex]\infty[/tex] = 2[tex]\infty[/tex]

But the only LR indeterminate for near that is 1[tex]\infty[/tex] type... I left it at that. I'm pretty sure my answer is incorrect. Anyone?
 
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You wrote cosx(1/x), pretty sure you meant cos(1/x)?

Anyways, I think you are correct, take ln of both sides. You now have

[tex]ln|y| = x \cdot ln \left(1 + cos \left( \frac{1}{x} \right) \right) = \frac{ln \left(1 + cos \left( \frac{1}{x} \right) \right)}{\frac{1}{x}}[/tex]

If you now evaluate the limit you get

[tex]\frac{ln(2)}{0}[/tex]

Which as you pointed out is not in the form 0/0, note that if you had [tex]1^{\infty}[/tex] then you would have had ln(1)/0 = 0/0
 

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