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

Nov 19, 2016

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

#### Explanation:

We don't have to expand these binomials. Factor out the greatest $x$ degree from the numerator and denominator.

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

Distributing the exponents:

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

As $x \rightarrow \infty$, both $\frac{1}{x} \rightarrow 0$ and $\frac{1}{x} ^ 2 \rightarrow 0$, since the denominator will be very large.

$= {\left(2 + 0\right)}^{4} / {\left(3 + 0\right)}^{2} = {2}^{4} / {3}^{2} = \frac{16}{9}$