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\[\lim_{x \rightarrow \infty}(\frac{ 5x-3 }{ 5x+7 })^{3x+1}\]

take the log
take the limit
exponentiate

I am allowed to use L'Hospital's rule. I am not sure how I would use it though...

So I get it's (infinity/infinity)^(infinity) but I am not sure how to go from there...

Yep... I am not sure how that would help though...

\[(3x+1)\ln(\frac{5x-3}{5x+7})\]
now you \(\infty\times 0\)

Yep... I got there.

Hmm Okay... Let's see...

probably easier if you write
\[\frac{\ln(5x-3)-\ln(5x+7)}{\frac{1}{3x+1}}\]

Okay Thanks!

Yep. I know.

yw

denominator is wrong

Chain rule. Right.

:( . Fair enough....

actually it is not that bad

Yep. I think I see why.

and don't forget to exponentiate at the end

Hmm... Okay.

So I got 1/e^6 or e^-6.

yup

:) . Thanks a million :) .