The discussion revolves around evaluating the limit of the difference of two natural logarithmic expressions as x approaches infinity: Limx→∞ Ln(x2 - 1) - Ln(2x2 + 3).
Participants are actively engaging with each other's reasoning, with some confirming the steps taken and others questioning the final interpretation of the limit. There is a recognition of the need to apply the logarithm to the limit of the fraction derived from the original expressions.
There appears to be some confusion regarding the treatment of the logarithmic terms and the final result, with participants suggesting different interpretations of the outcome.
What happened to the ##\ln##?AlexandraMarie112 said: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?
Yes that's what I did. So my final answer then should be Ln(1/2) ?Mastermind01 said: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.
Right, or -ln(2)AlexandraMarie112 said:Yes that's what I did. So my final answer then should be Ln(1/2) ?