Limit of $[\ln (3+x^2) - \ln (2+x)]$ as $x$ approaches $\infty$

Question

I got an answer as $\infty$ but I need more clear explanation.

Please help me!

Answer 1

Note that since

$$\log A-\log B= \log \frac{A}B$$

we have

$$\log (3+x^2) - \log (2+x)=\log\left(\frac{3+x^2}{2+x}\right)\to+\infty$$

indeed

$$\frac{3+x^2}{2+x}=x\frac{\frac3{x^2}+1}{\frac1{x}+1}\to +\infty$$

Answer 2

Hint Write $$ \lim_{x\to\infty}(\ln(3+x^2)-\ln(2+x))=\ln\left[\lim_{x\to\infty}\frac{3+x^2}{2+x}\right]. $$

Answer 3

Hint:

You can use equivalents, and the properties of log: $$\ln(3+x^2)-\ln(1+x)=\ln \frac{x^2+3}{x+1}.$$ On the other hand, $$\frac{x^2+3}{x+1}\sim_\infty \frac{x^2}x=\ln x\enspace\text{so}\quad\ln(3+x^2)-\ln(1+x)\sim_\infty \ln x.$$

Answer 4

$$\lim_{x \to \inf}ln(3+x^2) - ln(2+x)$$ can be symplified as $$\lim_{x \to \inf}ln( \frac{3+x^2}{2+x})$$ So, we can use this as a truth. $$\lim_{x \to \inf} \frac {1}{x} = 0$$ And we will have $$\lim_{x \to \inf}ln( \frac{x^2( \frac{3}{x^2}+1)}{x (\frac{2}{x}+1)})$$

As you see, you’ll have $x$ left so the answer is $\inf$