Limit in Infinity: Check Results Now

[Alexstrasuz1]

Screenshot by Lightshot
Screenshot by Lightshot
Can some one check results I got e^3/0 which is e^(infitnity) and for second I got e^(-5/2)

 

[Siron]

According to wolfram alpha you should get $e^3$ for the first limit and indeed $e^{\frac{-5}{2}}$ for the second one.

 

[Alexstrasuz1]

Oh yes it is 3 I made mistake with dividing, ty