Limit of Natural Log Sequence: How to Find It Using L'Hopital's Rule?

[tmt1]

I have this sequence:

$${a}_{n} = \ln \left(\frac{12n + 2}{-9 + 4n}\right)$$

I need to find the limit of this sequence. How can I go about this? Do I need to apply l'hospital's rule? I'm unsure how to simplify this expression. If I use the rule $\ln(\frac{a}{b}) = \ln a - \ln b$ I get $\infty - \infty$, which I don't think is useful.

 

[Sudharaka]

I have this sequence:

$${a}_{n} = \ln \left(\frac{12n + 2}{-9 + 4n}\right)$$

I need to find the limit of this sequence. How can I go about this? Do I need to apply l'hospital's rule? I'm unsure how to simplify this expression. If I use the rule $\ln(\frac{a}{b}) = \ln a - \ln b$ I get $\infty - \infty$, which I don't think is useful.

Hi tmt, :)

Here's a hint. Divide both the numerator and the denominator of the fraction inside the denominator by $n$ so that you get,

$${a}_{n} = \ln \left(\frac{12 + \frac{2}{n}}{-\frac{9}{n} + 4}\right)$$

Try to continue from here. :)