`lim_(x->oo) (sqrt(x^4)+4x^3)/(7x^2 + 1)`

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`lim_(x->oo) (sqrt(x^4)+4x^3)/(7x^2 + 1)`

Postby barnamah » Tue 12 April, 2011 18:51

What is `lim_(x->oo) (sqrt(x^4)+4x^3)/(7x^2 + 1)` ?
Explore and know. That is asked.
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Re: `lim_(x->oo) (sqrt(x^4)+4x^3)/(7x^2 + 1)`

Postby Integr88 » Tue 12 April, 2011 19:22

`lim_(x->oo) (sqrt(x^4)+4x^3)/(7x^2 + 1)`

`lim_(x->00) (sqrt(x^4)+4 x^3)/(14 x)`
Factor out constants:
`lim_(x->oo) (sqrt(x^4)+4 x^3)/(14 x)`
Factor out constants:
`= 1/14 (lim_(x->oo) (4 x^3+sqrt(x^4))/x)`

The limit of a sum is the sum of the limits:
`= 1/14 (lim_(x->oo) sqrt(x^4)/x+4 (lim_(x->oo) x^2))`

Using the power law, write` lim_(x->oo) x^2 as (lim_(x->oo) x)^2`:
`= 1/14 (lim_(x->oo) sqrt(x^4)/x+4 (lim_(x->oo) x)^2)`
The limit of x as x approaches `oo` is `oo`:
`= 1/14 (lim_(x->oo) sqrt(x^4)/x+oo)`
Simplify radicals, `sqrt(x^4)/x = sqrt(x^2)`:
`= 1/14 (lim_(x->oo) sqrt(x^2)+oo)`
Using the power law, write `lim_(x->oo) sqrt(x^2) as sqrt(lim_(x->oo) x^2)`:
`= 1/14 (sqrt(lim_(x->oo) x^2)+oo)`
Using the power law, write `lim_(x->oo) x^2 as (lim_(x->oo) x)^2`:
`= 1/14 (sqrt((lim_(x->oo) x)^2)+oo)`
The limit of x as x approaches `oo` is `oo`:
`= oo`
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