# Question #52e68

Sep 27, 2016

${\lim}_{x \to \infty} \ln \left({x}^{2} - 1\right) - \ln \left(2 {x}^{2} + 3\right) = \ln \left(\frac{1}{2}\right)$

#### Explanation:

First, we use the property of logarithms that ${\log}_{a} \left(x\right) - {\log}_{a} \left(y\right) = {\log}_{a} \left(\frac{x}{y}\right)$

${\lim}_{x \to \infty} \ln \left({x}^{2} - 1\right) - \ln \left(2 {x}^{2} + 3\right) = {\lim}_{x \to \infty} \ln \left(\frac{{x}^{2} - 1}{2 {x}^{2} - 3}\right)$

Next, we use the property of continuous functions that if $f \left(x\right)$ is continuous on an interval containing $a$ and ${\lim}_{x \to a} g \left(x\right)$ exists, then ${\lim}_{x \to a} f \left(g \left(x\right)\right) = f \left({\lim}_{x \to a} g \left(x\right)\right)$. As the natural log $\ln \left(x\right)$ is continuous on $\left(0 , \infty\right)$, that means

${\lim}_{x \to \infty} \ln \left(\frac{{x}^{2} - 1}{2 {x}^{2} - 3}\right) = \ln \left({\lim}_{x \to \infty} \frac{{x}^{2} - 1}{2 {x}^{2} - 3}\right)$

Now we just need to evaluate the limit of the rational expression within the logarithm.

${\lim}_{x \to \infty} \frac{{x}^{2} - 1}{2 {x}^{2} + 3} = {\lim}_{x \to \infty} \frac{1 - \frac{1}{x} ^ 2}{2 + \frac{3}{x} ^ 2}$

$= \frac{1 - \frac{1}{\infty}}{2 + \frac{3}{\infty}}$

$= \frac{1 - 0}{2 + 0}$

$= \frac{1}{2}$

Substituting that back into the logarithm, we get our final result:

${\lim}_{x \to \infty} \ln \left({x}^{2} - 1\right) - \ln \left(2 {x}^{2} + 3\right) = \ln \left({\lim}_{x \to \infty} \frac{{x}^{2} - 1}{2 {x}^{2} - 3}\right)$

$= \ln \left(\frac{1}{2}\right)$

(We could also use the property that $\log \left({a}^{x}\right) = x \log \left(a\right)$ to express the answer as $\ln \left(\frac{1}{2}\right) = \ln \left({2}^{- 1}\right) = - \ln \left(2\right)$)