Limit problem

  • Thread starter grothem
  • Start date
23
0
Lim x[tex]\rightarrow[/tex][tex]\infty[/tex] x^3 [tex]\times[/tex] e^(-x^2)

I moved x^3 down to make it a fraction and then I used L'Hospital's rule, but I can't come up with the right answer. Not sure what I'm doing wrong
 
23
0
that looks kind of confusing...

its the limit of x^3 * e^(-x^2) as x approaches infinity
 
1,631
4
well, i guess this is going to be zero, since [tex] e^{x^2}[/tex] grows much faster than x^3
 
206
0
What do you mean you "moved it down"?

Regardless, I suggest rewriting [itex]e^{-x^{2}}[/itex] using the facts you know about negative exponents. Probably easier that way.
 
1,631
4
and if you want to use l'hopitals rule, you need to do it three times from what i can see!
 
23
0
ah ok....I used l'hospital's rule 3 times and ended up with

lim (3/(4x^3*e^(x^2)) = 0
 
1,631
4
ah ok....I used l'hospital's rule 3 times and ended up with

lim (3/(4x^3*e^(x^2)) = 0
Well the bottom will defenitely not be what u got!
[e^(x^2)]'=(x^2)'e^(x^2)]=[2xe^(x^2)] not take again the derivative and apply the product rule

(uv)'=u'v+v'u
 

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