Question

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

Answer 1

`e^(-8)=0.0003355`, nearly. See the revelation that the limit is the same , as `x to - oo`.

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 oo`,

`lim ln y = lim (4x^2+2)ln((2x^2-1)/(2x^2+3))`

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

`= lim ((ln((2x^2-1)/(2x^2+3) ))') / ((1/(4x^2+2))',)` using L'Hospital rule

`= lim(4x/(2x^2-1)-4x/(2x^2+3))/((-1/(4x^2+2)^2)8x)`

`= lim -2(4x^2+2)^2 /((2x^2-1)(2x^2+3))`

`= lim -2((4+(2/x^2))^2)/((2-1/x^2)(2+3/x^2))`

`=-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 +-oo`.

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