Question

Find the limit as x approaches infinity of the function:

[(2x-3)/(2x+3)]^{2x+1}

I did this correctly before, but I suddenly forgot it. I would like some help here.

Answer 1

probably best bet is to take the log and compute the limit of that

Answer 2

So it is the fraction raised to the power (2x + 1)

Answer 3

Ok this is what I did:
Let y = that fcn
Then, ln(y) = ln(that fcn)
Now, I do not know what to do next.

Answer 4

pretty much when yo see the variable in the exponent, first step is to take the log

Answer 5

Oops, then also:
limit as x-> infinity ln(y) = limit as x-> infinity ln(that fcn)

Answer 6

Now I forgot what to do next.

Answer 7

\[(\frac{2x-3}{2x+3})^{2x+1} \] take the log get
\[(2x+1)\ln(\frac{2x-3}{2x+3})\]

Answer 8

now maybe expand or use l'hopital at this step

Answer 9

How would L'H even work? Wouldn't the inside of ln yield 2?

Answer 10

Sorry 0.

Answer 11

OHHHH

Answer 12

Right!

Answer 13

So you can move the limit into the ln argument right, because ln is continuous for x -> infinity?

Answer 14

And then you get ln(1), which simplifies to 0.

Answer 15

And then you move 1/(2x + 1) to the denominator.

Answer 16

Take the limit of that, which is also 0.

Answer 17

Got it.

Answer 18

you have
\[\infty\times 0\] as your form here, so you can write
\[(2x+1)\ln(\frac{2x-3}{2x+3})=\frac{\ln\frac{2x-3}{2x+3})}{\frac{1}{2x+1}}\]

Answer 19

Yes.

Answer 20

oh i am mistaken

Answer 21

very mistaken
this is a lot of work

Answer 22

maybe this would be easier
\[(2x+1)\left(\ln(2x-3)-\ln(2x+3)\right)\]

Answer 23

no, that is not easier

Answer 24

1/e^6

Answer 25

I guess I was wrong ....mathg8 is correct according to wolfram alpha

Answer 26

I got it !