How do I show that $\lim_{x\rightarrow\pm\infty}\frac{\ln(x^2+1)}{x} = 0$ without L'Hôspital's rule?

Question

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?

Answer 1

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.

Answer 2

rewrite your term in the form $\ln\left(\left(1+x^2)^{1/x^2}\right)\right)^x$

Answer 3

$\lim_{x\to \pm \infty} \frac{\ln(x^2+1)}{x} = \lim_{x \to \pm\infty}\ln\left((x^2+1)^{\large\frac{1}{x}}\right) = \ln \left(\lim_{x\to \pm \infty}(x^2+1)^{\large \frac{1}{x}}\right) = \ln \left(\lim_{x\to \pm \infty}(x^2*(1+1/x^2))^{\large \frac{1}{x}}\right) = \ln \left(\lim_{x\to \pm \infty}(x^{1/x})^2*(1+1/x^2)^{\large \frac{1}{x}}\right)$

$= \ln((1)^2 * (1+0)^0) = \ln1$

Answer 4

$f(x)=\frac{\log(1+x^2)}{x}$ is an odd function, hence we just need to compute the limit for $x\to +\infty$, the other one is just the opposite. We have: $$0\leq\frac{\log(1+x^2)}{x}\leq 2\frac{\log(x+1)}{x}\leq\frac{2}{\sqrt{x}}$$ since $\log(x+1)\leq\sqrt{x}$ over $\mathbb{R}^+$, hence both limits are zero by squeezing.

Answer 5

The limit at $\infty$ is the same as $$ \lim_{t\to0^+}t\log(1+1/t^2)= \lim_{t\to0^+}t(\log(1+t^2)-\log(t^2)) $$ Now, $\lim_{t\to0^+}t\log(t^2)=\lim_{t\to0^+}2t\log t =0$ (this should be already known), so you want to look at $$ \lim_{t\to0^+}t\log(1+t^2)=\lim_{t\to0^+}t^3\frac{\log(1+t^2)}{t^2} $$ Since $$ \lim_{t\to0^+}\frac{\log(1+t^2)}{t^2}= \lim_{u\to0^+}\frac{\log(1+u)}{u}=1 $$ (a well known limit), you finally get $$ \lim_{x\to\infty}\frac{\log(1+x^2)}{x}=0 $$ Since the function is odd, you also have $$ \lim_{x\to-\infty}\frac{\log(1+x^2)}{x}=0 $$