Find $\lim_{x \to 0} \frac{\ln (x^2+1)} {x^2} $ without L'hopital's rule

Question

I have to find the limit without L'hopital's rule: $$\lim_{x \to 0} \frac{\ln (x^2+1)} {x^2} $$

Is it possible? I thought about using squeeze theorem or something, but it didn't work out.

Hints are more than welcome!

P.S - I didn't study Taylor series or Integrals yet.

Answer 1

$$\begin{align} \lim_{x \to 0} \frac{\ln (x^2+1)} {x^2}&=\lim_{x \to 0} \ln (x^2+1)^{\frac{1}{x^2}}\\ &=\ln\left(\lim_{x \to 0} (x^2+1)^{\frac{1}{x^2}}\right)\\ &=\ln e=1 \end{align}$$

Answer 2

If you know about equivalents, you have

$$\ln(x+1)\underset{x\to 0}{\sim}x$$

so

$$\ln(x^2+1)\underset{x\to 0}{\sim}x^2.$$

Therefore,

$$\lim_{x\to 0} \frac{\ln(x^2+1)}{x^2}=\lim_{x\to0} \frac{x^2}{x^2}=1.$$

Answer 3

Let $f(u)=e^u$. With $t=\ln(x^2+1)$ we get

$$\lim_{x \to 0} \frac{\ln (x^2+1)} {x^2}=\lim_{t \to 0} \frac{t-0} {f(t)-f(0)}=\frac{1}{f'(0)}=1.$$

Answer 4

With series:

$ \ln(x^2+1)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-+...$ for $|x|<1$,

hence

$\frac{\ln (x^2+1)} {x^2}=1-\frac{x^2}{2}+\frac{x^4}{3}-+... \to 1$ for $x \to 0$