Limit in Infinity: Check Results Now

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SUMMARY

The discussion centers on evaluating limits involving exponential functions, specifically e^3/0 and e^(-5/2). The user initially miscalculated the first limit, mistakenly interpreting it as e^(infinity) instead of the correct result of e^3, as confirmed by Wolfram Alpha. The second limit was correctly identified as e^(-5/2). This highlights the importance of careful calculation in limit evaluation.

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Alexstrasuz1
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According to wolfram alpha you should get $e^3$ for the first limit and indeed $e^{\frac{-5}{2}}$ for the second one.
 
Oh yes it is 3 I made mistake with dividing, ty
 

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