Evaluate (pictured below)

• LammettHash: The exponential term dominates, so that the limit is 0.
  
  To see why: Suppose we take [tex]y=-x[/tex], so that as [tex]x\to-\infty[/tex] we have [tex]y\to+\infty[/tex]. Now
  
  [tex]\displaystyle\lim_{x\to-\infty}x^4e^x=\lim_{y\to+\infty}(-y)^4e^{-y}=\lim_{y\to+\infty}\frac{y^4}{e^y}[/tex]
  
  Now recall that for all [tex]y>0[/tex], we have [tex]y>\ln y[/tex], which means [tex]e^y>y[/tex]. We can similarly argue that for sufficiently large values of [tex]y[/tex], we have [tex]e^y>y^n[/tex] for all integers [tex]n[/tex]. So the denominator in the limit with respect to [tex]y[/tex] will always (eventually) exceed the numerator and make the entire expression approach 0.