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## Dido525 3 years ago Find the Limit:

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1. Dido525

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

2. satellite73

take the log take the limit exponentiate

3. Dido525

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

4. Dido525

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

5. Dido525

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

6. satellite73

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

7. Dido525

Yep... I got there.

8. satellite73

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

9. Dido525

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

10. satellite73

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

11. Dido525

Okay Thanks!

12. Dido525

Yep. I know.

13. satellite73

yw

14. Dido525

@satellite73 : Okay so I have: $\lim_{x \rightarrow \infty} \frac{ \frac{ 5 }{ 5x-3 }-\frac{ 5 }{ 5x+7 } }{ \frac{ -2 }{ (3x+1)^2 } }$

15. satellite73

denominator is wrong

16. satellite73

$\lim_{x \rightarrow \infty} \frac{ \frac{ 5 }{ 5x-3 }-\frac{ 5 }{ 5x+7 } }{ \frac{ -3 }{ (3x+1)^2 } }$

17. Dido525

Chain rule. Right.

18. satellite73

ugly retricealgebra from here on in have fun, but you will get a rational expression so limit is same as horizontal asymptote

19. Dido525

:( . Fair enough....

20. satellite73

actually it is not that bad

21. Dido525

Yep. I think I see why.

22. satellite73

$-\frac{(3x+1)^2}{3}\times \frac{50}{(5x-3)(5x+7)}$ and not need to multiply out, just look at the ratio of the leading coefficients

23. satellite73

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 $$-6$$

24. satellite73

and don't forget to exponentiate at the end

25. Dido525

Hmm... Okay.

26. Dido525

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

27. satellite73

yup

28. Dido525

:) . Thanks a million :) .

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