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

Dec 1, 2016

${\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2} = \frac{16}{9}$

#### Explanation:

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

${\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2} = {\lim}_{u \rightarrow \infty} {\left[u \left(2 + \frac{1}{u}\right)\right]}^{4} / {\left[{u}^{2} \left(3 + \frac{1}{u} ^ 2\right)\right]}^{2}$

$\textcolor{w h i t e}{{\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2}} = {\lim}_{u \rightarrow \infty} \frac{{u}^{4} {\left(2 + \frac{1}{u}\right)}^{4}}{{\left({u}^{2}\right)}^{2} {\left(3 + \frac{1}{u} ^ 2\right)}^{2}}$

$\textcolor{w h i t e}{{\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2}} = {\lim}_{u \rightarrow \infty} {\left(2 + \frac{1}{u}\right)}^{4} / {\left(3 + \frac{1}{u} ^ 2\right)}^{2}$

As $u \rightarrow \infty$, or as $u$ becomes increasingly large, we see that $\frac{1}{u}$ and $\frac{1}{u} ^ 2$ get smaller and smaller denominators up to the point where $\frac{1}{u} , \frac{1}{u} ^ 2 \rightarrow 0$.

$\textcolor{w h i t e}{{\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2}} = {\left(2\right)}^{4} / {\left(3\right)}^{2}$

$\textcolor{w h i t e}{{\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2}} = \frac{16}{9}$

Dec 2, 2016

${\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2} = \frac{16}{9}$

#### Explanation:

${\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2} = {\lim}_{u \rightarrow \infty} \frac{{\left(2 u\right)}^{4} + 4 {\left(2 u\right)}^{3} + 6 {\left(2 u\right)}^{2} + 4 \left(2 u\right) + 1}{{\left(3 {u}^{2}\right)}^{2} + 2 \left(3 {u}^{2}\right) + 1}$

$\therefore {\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2} = {\lim}_{u \rightarrow \infty} \frac{16 {u}^{4} + 32 {u}^{3} + 24 {u}^{2} + 8 u + 1}{9 {u}^{4} + 6 {u}^{2} + 1}$

$\therefore {\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2} = {\lim}_{u \rightarrow \infty} \frac{16 {u}^{4} + 32 {u}^{3} + 24 {u}^{2} + 8 u + 1}{9 {u}^{4} + 6 {u}^{2} + 1} \cdot \frac{\frac{1}{u} ^ 4}{\frac{1}{u} ^ 4}$

$\therefore {\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2} = {\lim}_{u \rightarrow \infty} \frac{16 + \frac{32}{u} + \frac{24}{u} ^ 2 + \frac{8}{u} ^ 3 + \frac{1}{u} ^ 4}{9 + \frac{6}{u} ^ 2 + \frac{1}{u} ^ 4}$
$\therefore {\lim}_{u \rightarrow \infty} {\left(2 u + 1\right)}^{4} / {\left(3 {u}^{2} + 1\right)}^{2} = \frac{16}{9}$

We can confirm this with a graph:

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