How do I show that $\lim_{x\rightarrow\pm\infty}\frac{\ln(x^2+1)}{x} = 0$ without L'Hôspital's rule? Without the use of series of L'Hôspital's rule, therefore with common limits and some intuition. I wanted to use the common limit saying $\lim_{x\rightarrow0}\dfrac{\ln(1+x)}{x} = 1$, but that doesn't work as the limit there goes to $0$ and not $1$. Is it even possible without making it difficult to compute this limit without L'Hôspitals or series?

Let me consider only the $x\to+\infty$ case, as the other case follows trivially. Substituting $x\mapsto e^x$ shows $$\lim_{x\to\infty}\frac{\ln x}x=\lim_{x\to\infty}\frac x{e^x}=0$$ Now $$0\le\frac{\ln(x^2+1)}x\le\frac{\ln(2x^2)}x=\frac{\ln 2+2\ln x}x\to 0$$ So the limit is zero by squeezing.