Question

Question #d3ab0

Answer 1

`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`

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

Because

• `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`

Answer 2

`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`

Answer 3

`-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`