lim x→0 (ln(a + x)+ln(a - x)-2 ln a)/x²

Explanation:

Step1: Use logarithm properties

First, use the properties of logarithms $\ln m+\ln n=\ln(mn)$ and $n\ln m = \ln(m^n)$. $\ln(a + x)+\ln(a - x)-2\ln a=\ln((a + x)(a - x))-\ln(a^{2})=\ln\left(\frac{a^{2}-x^{2}}{a^{2}}\right)=\ln\left(1-\frac{x^{2}}{a^{2}}\right)$ So the limit becomes $\lim_{x\rightarrow0}\frac{\ln\left(1-\frac{x^{2}}{a^{2}}\right)}{x^{2}}$.

Step2: Use the equivalent - infinitesimal substitution

When $u\rightarrow0$, $\ln(1 + u)\sim u$. Let $u =-\frac{x^{2}}{a^{2}}$, when $x\rightarrow0$, $u\rightarrow0$. So $\ln\left(1-\frac{x^{2}}{a^{2}}\right)\sim-\frac{x^{2}}{a^{2}}$ as $x\rightarrow0$. Then $\lim_{x\rightarrow0}\frac{\ln\left(1-\frac{x^{2}}{a^{2}}\right)}{x^{2}}=\lim_{x\rightarrow0}\frac{-\frac{x^{2}}{a^{2}}}{x^{2}}$.

Step3: Calculate the limit

$\lim_{x\rightarrow0}\frac{-\frac{x^{2}}{a^{2}}}{x^{2}}=-\frac{1}{a^{2}}$

Answer:

$-\frac{1}{a^{2}}$