Calculate $\lim_{x\to 0}\frac{\ln{(1+x^2)}-x^2}{x^4}$ How does one calculate the following limit? $$\lim_{x\to 0}\frac{\ln{(1+x^2)}-x^2}{x^4}$$
My initial thought was to write $\ln(1+x^2)$ as $\frac{\ln(1+x^2)}{x^2}x^2$ which is equal to $1$ when $x$ approaches $0$, but this is not allowed as the bottom function $x^4$ also approaches zero. Is applying L'Hopital multiple times the solution? Or am I missing a trick to make my work easier in this case?
 A: Yes you can apply l'Hospital, but once is enough.
First you can easily check that the conditions for l'Hospital are met, then differentiating the numerator and denominator once gives
$$
\frac{\frac{2x}{1+x^2}-2x}{4x^3}=\frac{2x-2x(1+x^2)}{4x^3(1+x^2)}=\frac{-2x^3}{4x^3(1+x^2)}\xrightarrow{x\to0}-\frac{1}{2}.
$$
Hence the limit is $-\frac{1}{2}$.
A: Substitute $t=x^2.$ The limit becomes
$$\lim_{t\to 0^+}{\ln(1+t)-t\over t^2}$$ By application l'Hopital rule we get $${{1\over 1+t}-1\over 2t}=-{1\over 2(1+t)}\underset{t\to 0^+}{\longrightarrow} -{1\over 2}$$
A: Knowing that $\log(1+t)\sim t-t^2/2$ when $x\to 0$, we have:
$$\frac{\log{(1+x^2)}-x^2}{x^4}\sim\frac{x^2-x^4/2-x^2}{x^4}\overset{x\to0}\longrightarrow-\frac{1}{2} $$
A: $$L=\lim_{x\to 0}\frac{\ln{(1+x^2)}-x^2}{x^4}$$
Let $y=x^4$
$$L=\lim_{y\to 0}\frac{\ln{(1+\sqrt{y})}-\sqrt{y}}{y}$$
With "L-H Rule" , we get :
$$L=\lim_{y\to 0}\frac{1/[{2\sqrt{y}(1+\sqrt{y})}]-1/[2\sqrt{y}]}{1}$$
$$L=\lim_{y\to 0}\frac{1-(1+\sqrt{y})}{{2\sqrt{y}(1+\sqrt{y})}}$$
$$L=\lim_{y\to 0}\frac{-\sqrt{y}}{{2\sqrt{y}(1+\sqrt{y})}}$$
$$L=\lim_{y\to 0}\frac{-1}{{2(1+\sqrt{y})}}$$
$$L=-1/2$$
