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

Answer 1

Aurora Alvarado

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

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]}

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Answer 2

Ezra Adams

The limit as x approaches infinity of ((2x^2-1)/(2x^2+3))^(4x^2+2) is equal to 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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