I need to evaluate the following limit :
$$\lim\limits_{x \to \infty} \frac{\ln(x^2+1)}{x^2}$$
Using L'Hospital rule, I get this result (which I'm pretty sure is good)
$$\lim\limits_{x \to \infty} \frac{\frac{2x}{x^2+1}}{2x}$$
Now, I'm not sure how I must evaluate this. Either as :
$$\lim\limits_{x \to \infty} \frac{2x}{x^2+1}*\frac{1}{2x} = \frac{1}{x^2+1}$$
or
$$\lim\limits_{x \to \infty} 2x*(\frac{x^2+1}{2x})^{-1} = \frac{4x^2}{x^2+1}$$
According to most of the tools I use to validate my maths, the 1st result is the good one, but I can't figure out why, am I simply missing a set of parenthesis in my equation?
This is just elementary algebra:
$${{a\over b}\over c} = {a\over b\cdot c}$$
i.e., $a/b$ divided by $c$ equals $a$ divided by $bc$. Thus, the first result which you have obtained is the correct one. In the second equation, you wrote
$$2x\cdot\left(\frac{x^2+1}{2x}\right)^{-1}$$ instead of the correct $${1\over 2x}\cdot\left(\frac{x^2+1}{2x}\right)^{-1}$$