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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...

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Other answers:

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.
rewrite as \[\frac{\ln(\frac{5x-3}{5x+7})}{\frac{1}{3x+1}}\] now you have \[\frac{0}{0}\] and now you can use l'hopital
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.
@satellite73 : Okay so I have: \[\lim_{x \rightarrow \infty} \frac{ \frac{ 5 }{ 5x-3 }-\frac{ 5 }{ 5x+7 } }{ \frac{ -2 }{ (3x+1)^2 } }\]
denominator is wrong
\[\lim_{x \rightarrow \infty} \frac{ \frac{ 5 }{ 5x-3 }-\frac{ 5 }{ 5x+7 } }{ \frac{ -3 }{ (3x+1)^2 } }\]
Chain rule. Right.
ugly retricealgebra from here on in have fun, but you will get a rational expression so limit is same as horizontal asymptote
:( . Fair enough....
actually it is not that bad
Yep. I think I see why.
\[-\frac{(3x+1)^2}{3}\times \frac{50}{(5x-3)(5x+7)}\] and not need to multiply out, just look at the ratio of the leading coefficients
i mean you have to be a bit careful but the leading coefficient of the numerator is \(9\times 50\) and the leading coefficient of the denominator is \(-3\times 25\) so you the ratio is \(-6\)
and don't forget to exponentiate at the end
Hmm... Okay.
So I got 1/e^6 or e^-6.
:) . Thanks a million :) .

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