Question

How do you find the limit of `(2u+1)^4/(3u^2+1)^2` as `u->oo`?

Answer 1

`lim_(urarroo)(2u+1)^4/(3u^2+1)^2=16/9`

Explanation:

The trick with these is to factor out the greatest degree of `u` that we can from the numerator and denominator:

`lim_(urarroo)(2u+1)^4/(3u^2+1)^2=lim_(urarroo)[u(2+1/u)]^4/[u^2(3+1/u^2)]^2`

`color(white)(lim_(urarroo)(2u+1)^4/(3u^2+1)^2)=lim_(urarroo)(u^4(2+1/u)^4)/((u^2)^2(3+1/u^2)^2)`

`color(white)(lim_(urarroo)(2u+1)^4/(3u^2+1)^2)=lim_(urarroo)(2+1/u)^4/(3+1/u^2)^2`

As `urarroo`, or as `u` becomes increasingly large, we see that `1/u` and `1/u^2` get smaller and smaller denominators up to the point where `1/u,1/u^2rarr0`.

`color(white)(lim_(urarroo)(2u+1)^4/(3u^2+1)^2)=(2)^4/(3)^2`

`color(white)(lim_(urarroo)(2u+1)^4/(3u^2+1)^2)=16/9`

Answer 2

`lim_(u rarr oo)(2u+1)^4/(3u^2+1)^2 = 16/9`

Explanation:

`lim_(u rarr oo)(2u+1)^4/(3u^2+1)^2 = lim_(u rarr oo) ((2u)^4 + 4(2u)^3 + 6(2u)^2+4(2u)+1)/((3u^2)^2+2(3u^2)+1)`

`:. lim_(u rarr oo)(2u+1)^4/(3u^2+1)^2 = lim_(u rarr oo) (16u^4 + 32u^3 + 24u^2+8u+1)/(9u^4+6u^2+1)`

`:. lim_(u rarr oo)(2u+1)^4/(3u^2+1)^2 = lim_(u rarr oo) (16u^4 + 32u^3 + 24u^2+8u+1)/(9u^4+6u^2+1) * (1/u^4)/(1/u^4)`

`:. lim_(u rarr oo)(2u+1)^4/(3u^2+1)^2 = lim_(u rarr oo) (16 + 32/u + 24/u^2+8/u^3+1/u^4)/(9+6/u^2+1/u^4)`
`:. lim_(u rarr oo)(2u+1)^4/(3u^2+1)^2 = 16/9`

We can confirm this with a graph:

graph{(2x+1)^4/(3x^2+1)^2 [-3.25, 16.75, -4.48, 5.52]}