Limit of $\ln⁡(1+2x^2 )/\ln⁡(x^4+3)$ I've tried finding the limit
$$\lim_{x \to \infty} \frac{\ln(1+2x^2)}{\ln(x^4+3)}$$
using L'Hôpital's rule and got $0$. My CAS says that the limit should be $\frac 12$. 
Can you please give me a hint on what to do? 
 A: I hope you mean $x\rightarrow\infty$.
$$\lim_{x\rightarrow\infty}\frac{{\mathrm{\ln} \left(1+2x^2\right)\ }}{{\mathrm{\ln} \left(x^4+3\right)\ }}=\lim_{x\rightarrow\infty}\frac{2\ln|x|+{\mathrm{\ln} \left(\frac{1}{x^2}+2\right)\ }}{4\ln|x|+{\mathrm{\ln} \left(1+\frac{3}{x^4}\right)\ }}=\frac{1}{2}$$
Actually, $x\rightarrow\infty$ it's $x\rightarrow\pm\infty$. 
A: Repeatedly rearrange and drop terms that become irrelevant in the limit:
$$\lim_{x\to\infty}\frac{\ln(2x^2 + 1)}{\ln(x^4+3)} = \lim_{x\to\infty}\frac{\ln(2x^2 )}{\ln(x^4)} = \lim_{x\to\infty}\frac{2\ln(x) + \ln(2)}{4\ln(x)} = \lim_{x\to\infty}\frac{2\ln(x)}{4\ln(x)} = \frac{1}{2}$$
A: $$\frac{\text{d}}{\text{d}x}[\ln(1+2x^2)]=\frac{\text{d}}{\text{d}x}[1+2x^2]\cdot\frac{1}{1+2x^2}=\frac{4x}{1+2x^2}$$
$$\frac{\text{d}}{\text{d}x}[\ln(x^4+3)]=\frac{\text{d}}{\text{d}x}[x^4+3]\cdot\frac{1}{x^4+3}=\frac{4x^3}{x^4+3}$$
Then $$\lim\limits_{x\rightarrow\infty}\frac{\ln(1+2x^2)}{\ln(x^4+3)}=\lim\limits_{x\rightarrow\infty}\frac{\frac{4x}{1+2x^2}}{\frac{4x^3}{x^4+3}}=\lim\limits_{x\rightarrow\infty}\frac{4x}{1+2x^2}\frac{x^4+3}{4x^3}=\lim\limits_{x\rightarrow\infty}\frac{4x}{4x^3}\frac{x^4+3}{1+2x^2}=\lim\limits_{x\rightarrow\infty}\frac{1}{x^2}\frac{x^4+3}{1+2x^2}=\lim\limits_{x\rightarrow\infty}\frac{x^4+3}{2x^4+x^2}=\lim\limits_{x\rightarrow\infty}\frac{1+\frac{3}{x^4}}{2+\frac{1}{x^2}}=\frac{\lim\limits_{x\rightarrow\infty}(1+\frac{3}{x^4})}{\lim\limits_{x\rightarrow\infty}(2+\frac{1}{x^2})}=\frac{1+\lim\limits_{x\rightarrow\infty}\frac{3}{x^4}}{2+\lim\limits_{x\rightarrow\infty}\frac{1}{x^2}}=\frac{1}{2}$$
(There are better ways to tackle this problem as shown by the other answers, but this is how you correctly do it with L'Hopital)
A: As often, using equivalents makes things simpler:


*

*$2x^2+1\sim_\infty 2x^2$, so $\;\ln(2x^2+1)\sim_\infty\ln(2x^2)=\ln 2+2\ln x\sim_\infty2\ln x$,

*$x^4+3\sim_\infty x^4$, so $\;\ln(x^4+3)\sim_\infty 4\ln x$,
and finally
$$\frac{\ln(2x^2+1)}{\ln(x^4+3)}\sim_\infty\frac{2\ln x}{4\ln x}=\frac12.$$

