The limit of the expression Limx→∞ Ln(x2-1) - Ln(2x2+3) simplifies to Ln(1/2) or equivalently -Ln(2). The correct approach involves applying the properties of logarithms to combine the terms into a single logarithmic expression. By dividing both the numerator and denominator by x2, the limit of the fraction inside the logarithm approaches 1/2 as x approaches infinity, leading to the final result.
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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) ?