Inifinity limit with natural log

[AlexandraMarie112]

Homework Statement

Limx--> ∞ Ln(x^2-1) -Ln(2x^2+3)

Homework Equations

The Attempt at a Solution

Ln(x^2-1)/(2x^2+3)

Then I divided the top and bottom by x^2 so in the end I got (1/2).

Is this right?

 

[LCKurtz]

Homework Statement

Limx--> ∞ Ln(x^2-1) -Ln(2x^2+3)

Homework Equations

The Attempt at a Solution

Ln(x^2-1)/(2x^2+3)

Then I divided the top and bottom by x^2 so in the end I got (1/2).

Is this right?

What happened to the $\ln$?

 

[Mastermind01]

Is this what you did? :

$\lim_{n\rightarrow +\infty} {\ln {(x^2 - 1)} - \ln{(2x^2+3)}}$

$= \lim_{n\rightarrow +\infty} {\ln ({\frac{x^2 - 1}{2x^2+3}})}$$= {\ln {\lim_{n\rightarrow +\infty}(\frac{1 - \frac{1}{x^2}}{2+\frac{3}{x^2}}})}$

You got the limit of the inside part as $\frac{1}{2}$ you need to take its $\ln$ to get the right answer.

 

[AlexandraMarie112]

Is this what you did? :

$\lim_{n\rightarrow +\infty} {\ln {(x^2 - 1)} - \ln{(2x^2+3)}}$

$= \lim_{n\rightarrow +\infty} {\ln ({\frac{x^2 - 1}{2x^2+3}})}$$= {\ln {\lim_{n\rightarrow +\infty}(\frac{1 - \frac{1}{x^2}}{2+\frac{3}{x^2}}})}$

You got the limit of the inside part as $\frac{1}{2}$ you need to take its $\ln$ to get the right answer.

Yes that's what I did. So my final answer then should be Ln(1/2) ?

 

[Mark44]

Yes that's what I did. So my final answer then should be Ln(1/2) ?

Right, or -ln(2)