L'Hopitals Rule, Limit, x->infty, {(15x-3)/(15x+5)}^(3x+1)

[ifailedcalculus]

Hi, First time poster.

I need help with one question that uses L'Hopitals rule

this is the question:

Evaluate the following limit
lim
x-->infinity {(15x-3)/(15x+5)}^(3x+1)

What I've done so far is get it to look like this

ln y = (3x+1)*ln(1-8/(15x+5))

I know you have to apply L'Hopitals rule, probably twice...I can't figure this out...please help!

Thanks

 

[ifailedcalculus]

Re: L'Hopitals Rule, Limit

Any help would be appreciated! I need to figure this out by tomorrow

 

[BigGlenntheHeavy]

Re: L'Hopitals Rule, Limit

Hint: ln(y) = (3x+1)ln|15x-3|-(3x+1)ln|15x+5|
x->infinity.

Note: We let y = f(x).

 

[ifailedcalculus]

Re: L'Hopitals Rule, Limit

sorry, I still have no idea how to solve this.

What i did was find the first derivative of the numerator and the denominator, seperately, and went on to find the second derivative. I can't get the calculations right and I'm pretty sure I'm doing it wrong.

 

[BigGlenntheHeavy]

Re: L'Hopitals Rule, Limit

lim[ln(y)] = (3x+1)ln(|15x-3|/|15x+5|) = ln(|15x-3|/|15x+5|)/(1/(3x+1)).
x->infinity
As x-> infinity, we have ln(1)/0 = 0/0. Now take the derivitives of the numerator and the denominator and you'll get

-8(3x+1)/(15x-3). Now the limit as x approaches infinity is -infinity/infinity. Again take derivitives and you'll get -24/15 = -8/5.

Hence(limit) ln(y) = -8/5, y = e^(-8/5)
x->infinity


Ergo lim[f(x)] = e^(-8/5)
x->infinity

This one is quite protracted, gasp.

 

[ifailedcalculus]

Re: L'Hopitals Rule, Limit

Thank you so much Glenn!

I got past the first derivative and got the answer you wrote, and also almost got the second derivative correctly. What threw me off was I forgot to pair my answer with the "e".

That was a dumb mistake on my part.

Again, thanks a lot for helping me out with this question

 

[galactus]

We can also do this without L'Hopital. Unless you are required to use it.

Rewrite \(\displaystyle \frac{3(5x-1)}{5(3x+1)}=1-\frac{8}{5(3x+1)}\)

\(\displaystyle \lim_{x\to \infty}\left(1-\frac{8}{5(3x+1)}\right)^{3x+1}\)

Let t=3x+1

\(\displaystyle \lim_{t\to \infty}\left(1-\frac{8}{5t}\right)^{t}\)

Note the famous limit for e, \(\displaystyle e=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{t}\)

Therefore, we get \(\displaystyle e^{\frac{-8}{5}}\) as the limit.