Boilanubjaini8f

2022-06-25

Evaluate
$\underset{x\to 0}{lim}\left(\frac{1}{\mathrm{ln}\left(1+x\right)}+\frac{1}{\mathrm{ln}\left(1-x\right)}\right)$

Josie123

Expert

What's wrong with L'Hopital's?
$\begin{array}{rl}\underset{x\to 0}{lim}\frac{1}{\mathrm{ln}\left(1+x\right)}+\frac{1}{\mathrm{ln}\left(1-x\right)}& =\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1-{x}^{2}\right)}{\mathrm{ln}\left(1+x\right)\mathrm{ln}\left(1-x\right)}\\ & =\underset{x\to 0}{lim}\frac{-2x/\left(1-{x}^{2}\right)}{\mathrm{ln}\left(1-x\right)/\left(1+x\right)-\mathrm{ln}\left(1+x\right)/\left(1-x\right)}\\ & =\underset{x\to 0}{lim}\frac{-2x}{\left(1-x\right)\mathrm{ln}\left(1-x\right)-\left(1+x\right)\mathrm{ln}\left(1+x\right)}\\ & =\underset{x\to 0}{lim}\frac{-2}{-1-\mathrm{ln}\left(1-x\right)-1-\mathrm{ln}\left(1+x\right)}\\ & =\frac{-2}{-2}\\ & =1.\end{array}$

skylsn

Expert

Since $\mathrm{log}\left(1+x\right)=x-\frac{{x}^{2}}{2}+O\left({x}^{3}\right)$, we have $\mathrm{log}\left(1-x\right)=-x-\frac{{x}^{2}}{2}+O\left({x}^{3}\right)$. Therefore,
$\begin{array}{rl}\frac{1}{\mathrm{log}\left(1+x\right)}+\frac{1}{\mathrm{log}\left(1-x\right)}& =\frac{1}{x\left(1-\frac{x}{2}+O\left({x}^{2}\right)\right)}+\frac{1}{-x\left(1+\frac{x}{2}+O\left({x}^{2}\right)\right)}\\ \text{(}\ast \text{)}& & =\frac{1+\frac{x}{2}+O\left({x}^{2}\right)}{x}-\frac{1-\frac{x}{2}+O\left({x}^{2}\right)}{x}& =\frac{x+O\left({x}^{2}\right)}{x}\end{array}$
where $\left(\ast \right)$ is because $\frac{1}{1+x}=1-x+O\left({x}^{2}\right)$
Therefore,
$\underset{x\to 0}{lim}\frac{1}{\mathrm{log}\left(1+x\right)}+\frac{1}{\mathrm{log}\left(1-x\right)}=1$

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