# lim_ (x -> oo) ((2x^2-1)/(2x^2+3))^(4x^2+2)= ?

Dec 3, 2016

${e}^{- 8} = 0.0003355$, nearly. See the revelation that the limit is the same , as $x \to - \infty$.

#### Explanation:

At the outset, I have to state that we can enhance the merit of our

answers, with a befitting graph that can always be inserted, using

Socratic graphic facility. I have done it here, for this limit problem.

Let y denote the given expression. Then,

as $x \to \infty$,

$\lim \ln y = \lim \left(4 {x}^{2} + 2\right) \ln \left(\frac{2 {x}^{2} - 1}{2 {x}^{2} + 3}\right)$

= lim ln((2x^2-1)/(2x^2+3) )/(1/(4x^2+2), in the form 0/0

$= \lim \frac{\left(\ln \left(\frac{2 {x}^{2} - 1}{2 {x}^{2} + 3}\right)\right) '}{\left(\frac{1}{4 {x}^{2} + 2}\right) ' ,}$ using L'Hospital rule

$= \lim \frac{4 \frac{x}{2 {x}^{2} - 1} - 4 \frac{x}{2 {x}^{2} + 3}}{\left(- \frac{1}{4 {x}^{2} + 2} ^ 2\right) 8 x}$

$= \lim - 2 {\left(4 {x}^{2} + 2\right)}^{2} / \left(\left(2 {x}^{2} - 1\right) \left(2 {x}^{2} + 3\right)\right)$

$= \lim - 2 \frac{{\left(4 + \left(\frac{2}{x} ^ 2\right)\right)}^{2}}{\left(2 - \frac{1}{x} ^ 2\right) \left(2 + \frac{3}{x} ^ 2\right)}$

$= - 8$

So, $\lim y = {e}^{- 8}$.

The function is an even function, with y-intercept = 1/9, for max y.

Also, $y > {e}^{- 8} > 0$.

So, $y \to {e}^{- 8}$, as $x \to \pm \infty$.

graph{y-((2x^2+1)/(2x^2+3))^(4x^2+2)=0 [-2.5, 2.5, -1.25, 1.25]}