Question

1. $lim_{x

ightarrow0^{+}}\frac{ln(1 + x^{2})}{x^{3}}$

Explanation:

Step1: Use equivalent - infinitesimal

When $x\to0$, $\ln(1 + u)\sim u$. Here $u = x^{2}$, so $\ln(1 + x^{2})\sim x^{2}$ as $x\to0$.
$$\lim_{x\to0^{+}}\frac{\ln(1 + x^{2})}{x^{3}}=\lim_{x\to0^{+}}\frac{x^{2}}{x^{3}}$$

Step2: Simplify the limit

Simplify $\frac{x^{2}}{x^{3}}=\frac{1}{x}$.
$$\lim_{x\to0^{+}}\frac{1}{x}=+\infty$$

Answer:

$+\infty$