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