Question

another limit

Answer 1

\[\large \lim_{x \to \infty}(\frac{2x-3}{2x+5})^{2x+1}\]

Answer 2

I said let y = f(x)^{g(x)}

got lny = g(x)ln(fx)

Answer 3

simplified that into \[\lim_{x \to \infty} lny = lin_{x \to \infty}(2x+1)[\ln(2x-3)-\ln(2x+5)]\] now I am stuck because the ln(-3) is undefined...

Answer 4

i wouldn't do it that way i don't think
makes it more complicated to use the "difference of the logs" leave it as you had it

Answer 5

\[\large \lim_{x \to \infty}(2x+1)\ln(\frac{2x-3}{2x+5})\]

Answer 6

now it looks like \(\infty\times 0\) right?

Answer 7

o ok so \[\lim_{x \to\infty} (2x+1)[\ln \frac{2x-3}{2x+5}]\]back to the way I had it.

Answer 8

I need to change I to 0/0 or infinity over infinity
\[\LARGE \frac{\ln \frac{2x-3}{2x+5}}{\frac{1}{2x+1}}\]

Answer 9

that will work nicely i think
although the algebra is going to be ugly either way

Answer 10

oh well... alright lets see\[\frac{\frac{1}{\frac{2x-3}{2x+5}}*\frac{[(2)(2x+5)-(2x-3)(x)]}{(2x+5)^{2}}}{\frac{-2x}{(2x+1)^{2}}}\]

Answer 11

yeah it is ugly enough, although the derivatives are easy
now it may be time to break up the log, so that the derivative of the numerator is
\[\frac{2}{2x-3}-\frac{2}{2x+5}=\frac{16}{4x^2+4x-15}\]

Answer 12

and the derivative of the denominator is
\[-\frac{2}{(2x+1)^2}\] i think you had a typo there

Answer 13

oh I see that's much easier ..

Answer 14

yea

Answer 15

than do we need to simplify or can we just plug in infinity now

Answer 16

now invert and multiply to get
\[\frac{-16(2x+1)^2}{2(4x^2+4x-15)}\]

Answer 17

should we just divide bottom and top by 1/x^2

Answer 18

i am doing this one the fly, so check my algebra, i just inverted and multiplied, i think it is right

Answer 19

yea so do i

Answer 20

now do it with your eyeballs, as the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients

Answer 21

I got the form \[\frac{-16*4}{2*4}\]

Answer 22

in my head i get \(-8\) but again your should check

Answer 23

yeah, what you got

Answer 24

so the answer is \[e^{-8}\]

Answer 25

assuming we did not screw up the algebra, yes

Answer 26

yea I pretty sure you distributed correctly.

Answer 27

tada!!
http://www.wolframalpha.com/input/?i= \lim_{x+\to+\infty}+%28\frac{2x-3}{2x%2B5}%29^{2x%2B1}

Answer 28

O what it was 1/e^8 yea were your right XD