Question #d3ab0

3 Answers
Jan 2, 2018

#lim_(xrarr-oo)(4^x+2xe^x)/(4^x+x^2*5^x)=-oo#

Explanation:

#lim_(xrarr-oo)(4^x+2xe^x)/(4^x+x^2*5^x)#

#x->-oo# so we divide with #:4^x#

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#=# #lim_(xrarr-oo)(1+2(xe^x)/4^x)/(1+(x^2*5^x)/4^x)=-oo#

Because

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  • #lim_(xrarr-oo)(xe^x)/4^x=lim_(xrarr-oo)x(e/4)^x=-oo#

  • #lim_(xrarr-oo)(x^2*5^x)/4^x=lim_(xrarr-oo)x^2/(4/5)^x=#

#lim_(xrarr-oo)(2x)/((4/5)^x*ln(4/5))=2/ln(4/5)lim_(xrarr-oo)x/(4/5)^x=#

#2/ln(4/5)lim_(xrarr-oo)1/((4/5)^xln(4/5))=2/ln^2(4/5)lim_(xrarr-oo)1/(4/5)^x#

#=2/ln^2(4/5)*0=0#

Used Rules De L'Hospital for #lim_(xrarr-oo)x/(4/5)^x# &
#lim_(xrarr-oo)x^2/(4/5)^x#

Jan 2, 2018

#Lim_(xrarroo)(4^x+2xe^x)/(4^x+x^2*5^x)=0#

Explanation:

#(4^x+2xe^x)/(4^x+x^2*5^x)#

#2xe^x# increases "faster" than #4^x# so we focuse on that

#(2e^x)/(x*5^x)=2/x*(e/5)^x#

#(e/5)^x~~(0.5)^x# so when #xrarrooquadquadquad(0.5)^oo=0#

#=>2/oo*(e/5)^oo=0*0=0#

#Lim_(xrarroo)(4^x+2xe^x)/(4^x+x^2*5^x)=0#

Jan 2, 2018

#-oo#

Explanation:

#lim_(x->-oo)(4^x + 2 x e^x)/(4^x + x^2 5^x) = lim_(x->oo)(4^-x - 2 x e^-x)/(4^-x + x^2 5^-x) =#

Now

#(4^-x - 2 x e^-x)/(4^-x + x^2 5^-x) =(5^x/5^x)((4^-x - 2 x e^-x)/(4^-x + x^2 5^-x)) = ((5/4)^x-2x(5/e)^x)/((5/4)^x+x^2) = (1/x^2(5/4)^x-2/x(5/e)^x)/(1/x^2(5/4)^x+1) # but

#lim_(x->oo)(1/x^2(5/4)^x-2/x(5/e)^x)/(1/x^2(5/4)^x+1) =lim_(x->oo)(1/x^2(5/4)^x-2/x(5/e)^x)/(1/x^2(5/4)^x) = #
#=lim_(x->oo)1-2x(4/e)^x = -oo#

NOTE

Here #4/e > 1#